Sql queries interview questions

SQL Queries Interview Questions - Oracle Part 1
As a database developer, writing SQL queries, PLSQL code is part of daily life. Having a good
knowledge on SQL is really important. Here i am posting some practical exampleson SQL queries.

To solve these interview questions on SQL queries you have to create the products, sales tables in your
oracle database. The "Create Table", "Insert" statements are provided below.

CREATE TABLE PRODUCTS

(

PRODUCT_ID INTEGER,

PRODUCT_NAME VARCHAR2(30)

);

CREATE TABLE SALES

(

SALE_ID INTEGER,

PRODUCT_ID INTEGER,

YEAR INTEGER,

Quantity INTEGER,

PRICE INTEGER

);

INSERT INTO PRODUCTS VALUES ( 100, 'Nokia');

INSERT INTO PRODUCTS VALUES ( 200, 'IPhone');

INSERT INTO PRODUCTS VALUES ( 300, 'Samsung');

INSERT INTO PRODUCTS VALUES ( 400, 'LG');

INSERT INTO SALES VALUES ( 1, 100, 2010, 25, 5000);









COMMIT;

The products table contains the below data.

SELECT * FROM PRODUCTS;

PRODUCT_ID PRODUCT_NAME

-----------------------

100 Nokia

200 IPhone

300 Samsung

The sales table contains the following data.

SELECT * FROM SALES;

SALE_ID PRODUCT_ID YEAR QUANTITY PRICE

--------------------------------------

1 100 2010 25 5000

2 100 2011 16 5000

3 100 2012 8 5000

4 200 2010 10 9000

5 200 2011 15 9000

6 200 2012 20 9000

7 300 2010 20 7000

8 300 2011 18 7000

9 300 2012 20 7000

Here Quantity is the number of products sold in each year. Price is the sale price of each product.

I hope you have created the tables in your oracle database. Now try to solve the belowSQL queries.

1. Write a SQL query to find the products which have continuous increase in sales every year?

Solution:

Here “Iphone” is the only product whose sales are increasing every year.

STEP1: First we will get the previous year sales for each product. The SQL query to do this is

SELECT P.PRODUCT_NAME,

S.YEAR,

S.QUANTITY,

LEAD(S.QUANTITY,1,0) OVER (

PARTITION BY P.PRODUCT_ID

ORDER BY S.YEAR DESC

) QUAN_PREV_YEAR

FROM PRODUCTS P,

SALES S

WHERE P.PRODUCT_ID = S.PRODUCT_ID;

PRODUCT_NAME YEAR QUANTITY QUAN_PREV_YEAR

-----------------------------------------

Nokia 2012 8 16

Nokia 2011 16 25

Nokia 2010 25 0

IPhone 2012 20 15

IPhone 2011 15 10

IPhone 2010 10 0

Samsung 2012 20 18

Samsung 2011 18 20

Samsung 2010 20 0

Here the lead analytic function will get the quantity of a product in its previous year.

STEP2: We will find the difference between the quantities of a product with its previous year’s quantity.
If this difference is greater than or equal to zero for all the rows, then the product is a constantly
increasing in sales. The final query to get the required result is

SELECT PRODUCT_NAME

FROM

(


S.QUANTITY -

LEAD(S.QUANTITY,1,0) OVER (

PARTITION BY P.PRODUCT_ID

ORDER BY S.YEAR DESC

) QUAN_DIFF

FROM PRODUCTS P,

SALES S

WHERE P.PRODUCT_ID = S.PRODUCT_ID

)A

GROUP BY PRODUCT_NAME

HAVING MIN(QUAN_DIFF) >= 0;

PRODUCT_NAME

------------

IPhone

2. Write a SQL query to find the products which does not have sales at all?

Solution:

“LG” is the only product which does not have sales at all. This can be achieved in three ways.

Method1: Using left outer join.

SELECT P.PRODUCT_NAME

FROM PRODUCTS P

LEFT OUTER JOIN

SALES S

ON (P.PRODUCT_ID = S.PRODUCT_ID);

WHERE S.QUANTITY IS NULL

PRODUCT_NAME

------------

LG

Method2: Using the NOT IN operator.


FROM PRODUCTS P

WHERE P.PRODUCT_ID NOT IN

(SELECT DISTINCT PRODUCT_ID FROM SALES);

PRODUCT_NAME

------------

LG

Method3: Using the NOT EXISTS operator.


FROM PRODUCTS P

WHERE NOT EXISTS

(SELECT 1 FROM SALES S WHERE S.PRODUCT_ID = P.PRODUCT_ID);

PRODUCT_NAME

------------

LG

3. Write a SQL query to find the products whose sales decreased in 2012 compared to 2011?

Solution:

Here Nokia is the only product whose sales decreased in year 2012 when compared withthe sales in
the year 2011. The SQL query to get the required output is


FROM PRODUCTS P,

SALES S_2012,

SALES S_2011

WHERE P.PRODUCT_ID = S_2012.PRODUCT_ID

AND S_2012.YEAR = 2012

AND S_2011.YEAR = 2011

AND S_2012.PRODUCT_ID = S_2011.PRODUCT_ID

AND S_2012.QUANTITY < S_2011.QUANTITY;

PRODUCT_NAME

------------

Nokia

4. Write a query to select the top product sold in each year?

Solution:

Nokia is the top product sold in the year 2010. Similarly, Samsung in 2011 and IPhone, Samsung in
2012. The query for this is

SELECT PRODUCT_NAME,

YEAR

FROM

(


S.YEAR,

RANK() OVER (

PARTITION BY S.YEAR

ORDER BY S.QUANTITY DESC

) RNK

FROM PRODUCTS P,

SALES S


)A

WHERE RNK = 1;

PRODUCT_NAME YEAR

--------------------

Nokia 2010

Samsung 2011

IPhone 2012

Samsung 2012

5. Write a query to find the total sales of each product.?

Solution:

This is a simple query. You just need to group by the data on PRODUCT_NAME and then find the sum
of sales.


NVL( SUM( S.QUANTITY*S.PRICE ), 0) TOTAL_SALES

FROM PRODUCTS P

LEFT OUTER JOIN

SALES S

ON (P.PRODUCT_ID = S.PRODUCT_ID)

GROUP BY P.PRODUCT_NAME;

PRODUCT_NAME TOTAL_SALES

---------------------------

LG 0

IPhone 405000

Samsung 406000

Nokia 245000

This is continuation to my previous post, SQL Queries Interview Questions - Oracle Part 1, Where i
have used PRODUCTS and SALES tables as an example. Here also i am using the same tables. So,
just take a look at the tables by going through that link and it will be easy for you to understand
the questions mentioned here.

Solve the below examples by writing SQL queries.

1. Write a query to find the products whose quantity sold in a year should be greater than the average
quantity sold across all the years?

Solution:

This can be solved with the help of correlated query. The SQL query for this is


S.YEAR,

S.QUANTITY

FROM PRODUCTS P,

SALES S


AND S.QUANTITY >

(SELECT AVG(QUANTITY)

FROM SALES S1

WHERE S1.PRODUCT_ID = S.PRODUCT_ID

);

PRODUCT_NAME YEAR QUANTITY

--------------------------

Nokia 2010 25

IPhone 2012 20

Samsung 2012 20

Samsung 2010 20

2. Write a query to compare the products sales of "IPhone" and "Samsung" in each year? The output
should look like as

YEAR IPHONE_QUANT SAM_QUANT IPHONE_PRICE SAM_PRICE

---------------------------------------------------

2010 10 20 9000 7000

2011 15 18 9000 7000

2012 20 20 9000 7000

Solution:

By using self-join SQL query we can get the required result. The required SQL query is

SELECT S_I.YEAR,

S_I.QUANTITY IPHONE_QUANT,

S_S.QUANTITY SAM_QUANT,

S_I.PRICE IPHONE_PRICE,

S_S.PRICE SAM_PRICE

FROM PRODUCTS P_I,

SALES S_I,

PRODUCTS P_S,

SALES S_S

WHERE P_I.PRODUCT_ID = S_I.PRODUCT_ID

AND P_S.PRODUCT_ID = S_S.PRODUCT_ID

AND P_I.PRODUCT_NAME = 'IPhone'

AND P_S.PRODUCT_NAME = 'Samsung'

AND S_I.YEAR = S_S.YEAR

3. Write a query to find the ratios of the sales of a product?

Solution:

The ratio of a product is calculated as the total sales price in a particular year divide by the total
sales price across all years. Oracle provides RATIO_TO_REPORT analytical function for finding the
ratios. The SQL query is


S.YEAR,

RATIO_TO_REPORT(S.QUANTITY*S.PRICE)

OVER(PARTITION BY P.PRODUCT_NAME ) SALES_RATIO

FROM PRODUCTS P,

SALES S

WHERE (P.PRODUCT_ID = S.PRODUCT_ID);

PRODUCT_NAME YEAR RATIO

-----------------------------

IPhone 2011 0.333333333

IPhone 2012 0.444444444

IPhone 2010 0.222222222

Nokia 2012 0.163265306

Nokia 2011 0.326530612

Nokia 2010 0.510204082

Samsung 2010 0.344827586

Samsung 2012 0.344827586

Samsung 2011 0.310344828

4. In the SALES table quantity of each product is stored in rows for every year. Now write a query to
transpose the quantity for each product and display it in columns? The output should look like as

PRODUCT_NAME QUAN_2010 QUAN_2011 QUAN_2012

------------------------------------------

IPhone 10 15 20

Samsung 20 18 20

Nokia 25 16 8

Solution:

Oracle 11g provides a pivot function to transpose the row data into column data. The SQL query for this
is

SELECT * FROM

(


S.QUANTITY,

S.YEAR

FROM PRODUCTS P,

SALES S

WHERE (P.PRODUCT_ID = S.PRODUCT_ID)

)A

PIVOT ( MAX(QUANTITY) AS QUAN FOR (YEAR) IN (2010,2011,2012));

If you are not running oracle 11g database, then use the below query for transposing the row data into
column data.


MAX(DECODE(S.YEAR,2010, S.QUANTITY)) QUAN_2010,

MAX(DECODE(S.YEAR,2011, S.QUANTITY)) QUAN_2011,

MAX(DECODE(S.YEAR,2012, S.QUANTITY)) QUAN_2012

FROM PRODUCTS P,

SALES S

WHERE (P.PRODUCT_ID = S.PRODUCT_ID)

GROUP BY P.PRODUCT_NAME;

5. Write a query to find the number of products sold in each year?

Solution:

To get this result we have to group by on year and the find the count. The SQL query for this question is

SELECT YEAR,

COUNT(1) NUM_PRODUCTS

FROM SALES

GROUP BY YEAR;

YEAR NUM_PRODUCTS

------------------

2010 3

2011 3

2012 3

Here I am providing Oracle SQL Query Interview Questions. If you find any bugs in the queries, Please
do comment. So, that i will rectify them.

1. Write a query to generate sequence numbers from 1 to the specified number N?

Solution:

SELECT LEVEL FROM DUAL CONNECT BY LEVEL<=&N;

2. Write a query to display only friday dates from Jan, 2000 to till now?

Solution:

SELECT C_DATE,

TO_CHAR(C_DATE,'DY')

FROM

(

SELECT TO_DATE('01-JAN-2000','DD-MON-YYYY')+LEVEL-1 C_DATE

FROM DUAL

CONNECT BY LEVEL <=

(SYSDATE - TO_DATE('01-JAN-2000','DD-MON-YYYY')+1)

)

WHERE TO_CHAR(C_DATE,'DY') = 'FRI';

3. Write a query to duplicate each row based on the value in the repeat column? The input table data
looks like as below

Products, Repeat

----------------

A, 3

B, 5

C, 2

Now in the output data, the product A should be repeated 3 times, B should be repeated 5 times and C
should be repeated 2 times. The output will look like as below

Products, Repeat

----------------

A, 3

A, 3

A, 3

B, 5

B, 5

B, 5

B, 5

B, 5

C, 2

C, 2

Solution:

SELECT PRODUCTS,

REPEAT

FROM T,

( SELECT LEVEL L FROM DUAL

CONNECT BY LEVEL <= (SELECT MAX(REPEAT) FROM T)

)A

WHERE T.REPEAT >= A.L

ORDER BY T.PRODUCTS;

4. Write a query to display each letter of the word "SMILE" in a separate row?

S

M

I

L

E

Solution:

SELECT SUBSTR('SMILE',LEVEL,1) A

FROM DUAL

CONNECT BY LEVEL <=LENGTH('SMILE');

5. Convert the string "SMILE" to Ascii values? The output should look like as 83,77,73,76,69. Where 83
is the ascii value of S and so on.
The ASCII function will give ascii value for only one character. If you pass a string to the ascii function, it
will give the ascii value of first letter in the string. Here i am providing two solutions to get the ascii
values of string.

Solution1:

SELECT SUBSTR(DUMP('SMILE'),15)

FROM DUAL;

Solution2:

SELECT WM_CONCAT(A)

FROM

(

SELECT ASCII(SUBSTR('SMILE',LEVEL,1)) A

FROM DUAL

CONNECT BY LEVEL <=LENGTH('SMILE')

);

1. Consider the following friends table as the source

Name, Friend_Name

-----------------

sam, ram

sam, vamsi

vamsi, ram

vamsi, jhon

ram, vijay

ram, anand

Here ram and vamsi are friends of sam; ram and jhon are friends of vamsi and so on. Now write a
query to find friends of friends of sam. For sam; ram,jhon,vijay and anand are friends of friends. The
output should look as

Name, Friend_of_Firend

----------------------

sam, ram

sam, jhon

sam, vijay

sam, anand

Solution:

SELECT f1.name,

f2.friend_name as friend_of_friend

FROM friends f1,

friends f2

WHERE f1.name = 'sam'

AND f1.friend_name = f2.name;

2. This is an extension to the problem 1. In the output, you can see ram is displayed as friends of
friends. This is because, ram is mutual friend of sam and vamsi. Now extend the above query to
exclude mutual friends. The outuput should look as

Name, Friend_of_Friend

----------------------

sam, jhon

sam, vijay

sam, anand

Solution:

SELECT f1.name,

f2.friend_name as friend_of_friend

FROM friends f1,

friends f2

WHERE f1.name = 'sam'

AND f1.friend_name = f2.name

AND NOT EXISTS

(SELECT 1 FROM friends f3

WHERE f3.name = f1.name

AND f3.friend_name = f2.friend_name);

3. Write a query to get the top 5 products based on the quantity sold without using the row_number
analytical function? The source data looks as

Products, quantity_sold, year

-----------------------------

A, 200, 2009

B, 155, 2009

C, 455, 2009

D, 620, 2009

E, 135, 2009

F, 390, 2009

G, 999, 2010

H, 810, 2010

I, 910, 2010

J, 109, 2010

L, 260, 2010

M, 580, 2010

Solution:

SELECT products,

quantity_sold,

year

FROM

(

SELECT products,

quantity_sold,

year,

rownum r

from t

ORDER BY quantity_sold DESC

)A

WHERE r <= 5;

4. This is an extension to the problem 3. Write a query to produce the same output using row_number
analytical function?

Solution:

SELECT products,

quantity_sold,

year

FROM

(

SELECT products,

quantity_sold,

year,

row_number() OVER(

ORDER BY quantity_sold DESC) r

from t

)A

WHERE r <= 5;

5. This is an extension to the problem 3. write a query to get the top 5 products in eachyear based on
the quantity sold?

Solution:

SELECT products,

quantity_sold,

year

FROM

(

SELECT products,

quantity_sold,

year,

row_number() OVER(

PARTITION BY year

ORDER BY quantity_sold DESC) r

from t

)A

WHERE r <= 5;

SQL Query Interview Questions - Part 5
Write SQL queries for the below interview questions:

1. Load the below products table into the target table.

CREATE TABLE PRODUCTS

(

PRODUCT_ID INTEGER,

PRODUCT_NAME VARCHAR2(30)

);

INSERT INTO PRODUCTS VALUES ( 100, 'Nokia');

INSERT INTO PRODUCTS VALUES ( 200, 'IPhone');

INSERT INTO PRODUCTS VALUES ( 300, 'Samsung');

INSERT INTO PRODUCTS VALUES ( 400, 'LG');

INSERT INTO PRODUCTS VALUES ( 500, 'BlackBerry');

INSERT INTO PRODUCTS VALUES ( 600, 'Motorola');

COMMIT;

SELECT * FROM PRODUCTS;

PRODUCT_ID PRODUCT_NAME

-----------------------

100 Nokia

200 IPhone

300 Samsung

400 LG

500 BlackBerry

600 Motorola

The requirements for loading the target table are:

• Select only 2 products randomly.

• Do not select the products which are already loaded in the target table with in the last 30 days.

• Target table should always contain the products loaded in 30 days. It shouldnot contain the
products which are loaded prior to 30 days.

Solution:

First we will create a target table. The target table will have an additional column INSERT_DATE to
know when a product is loaded into the target table. The target
table structure is

CREATE TABLE TGT_PRODUCTS

(

PRODUCT_ID INTEGER,

PRODUCT_NAME VARCHAR2(30),

INSERT_DATE DATE

);

The next step is to pick 5 products randomly and then load into target table. While selecting check
whether the products are there in the

INSERT INTO TGT_PRODUCTS

SELECT PRODUCT_ID,

PRODUCT_NAME,

SYSDATE INSERT_DATE

FROM

(

SELECT PRODUCT_ID,

PRODUCT_NAME

FROM PRODUCTS S

WHERE NOT EXISTS (

SELECT 1

FROM TGT_PRODUCTS T

WHERE T.PRODUCT_ID = S.PRODUCT_ID

)

ORDER BY DBMS_RANDOM.VALUE --Random number generator in oracle.

)A

WHERE ROWNUM <= 2;

The last step is to delete the products from the table which are loaded 30 days back.

DELETE FROM TGT_PRODUCTS

WHERE INSERT_DATE < SYSDATE - 30;

2. Load the below CONTENTS table into the target table.

CREATE TABLE CONTENTS

(

CONTENT_ID INTEGER,

CONTENT_TYPE VARCHAR2(30)

);

INSERT INTO CONTENTS VALUES (1,'MOVIE');

INSERT INTO CONTENTS VALUES (2,'MOVIE');

INSERT INTO CONTENTS VALUES (3,'AUDIO');

INSERT INTO CONTENTS VALUES (4,'AUDIO');

INSERT INTO CONTENTS VALUES (5,'MAGAZINE');

INSERT INTO CONTENTS VALUES (6,'MAGAZINE');

COMMIT;

SELECT * FROM CONTENTS;

CONTENT_ID CONTENT_TYPE

-----------------------

1 MOVIE

2 MOVIE

3 AUDIO

4 AUDIO

5 MAGAZINE

6 MAGAZINE

The requirements to load the target table are:

• Load only one content type at a time into the target table.

• The target table should always contain only one contain type.

• The loading of content types should follow round-robin style. First MOVIE, second AUDIO, Third
MAGAZINE and again fourth Movie.

Solution:

First we will create a lookup table where we mention the priorities for the content types. The
lookup table “Create Statement” and data is shown below.

CREATE TABLE CONTENTS_LKP

(

CONTENT_TYPE VARCHAR2(30),

PRIORITY INTEGER,

LOAD_FLAG INTEGER

);

INSERT INTO CONTENTS_LKP VALUES('MOVIE',1,1);

INSERT INTO CONTENTS_LKP VALUES('AUDIO',2,0);

INSERT INTO CONTENTS_LKP VALUES('MAGAZINE',3,0);

COMMIT;

SELECT * FROM CONTENTS_LKP;

CONTENT_TYPE PRIORITY LOAD_FLAG

---------------------------------

MOVIE 1 1

AUDIO 2 0

MAGAZINE 3 0

Here if LOAD_FLAG is 1, then it indicates which content type needs to be loaded into the target table.
Only one content type will have LOAD_FLAG as 1. The other content types will have LOAD_FLAG as
0. The target table structure is same as the source tablestructure.

The second step is to truncate the target table before loading the data

TRUNCATE TABLE TGT_CONTENTS;

The third step is to choose the appropriate content type from the lookup table to load the source data
into the target table.

INSERT INTO TGT_CONTENTS

SELECT CONTENT_ID,

CONTENT_TYPE

FROM CONTENTS

WHERE CONTENT_TYPE = (SELECT CONTENT_TYPE FROM CONTENTS_LKP WHERE
LOAD_FLAG=1);

The last step is to update the LOAD_FLAG of the Lookup table.

UPDATE CONTENTS_LKP

SET LOAD_FLAG = 0

WHERE LOAD_FLAG = 1;

UPDATE CONTENTS_LKP

SET LOAD_FLAG = 1

WHERE PRIORITY = (

SELECT DECODE( PRIORITY,(SELECT MAX(PRIORITY) FROM CONTENTS_LKP) ,1 , PRIORITY+1)

FROM CONTENTS_LKP

WHERE CONTENT_TYPE = (SELECT DISTINCT CONTENT_TYPE FROM TGT_CONTENTS)

);

Grep Command in Unix and Linux Examples
Grep is the frequently used command in Unix (or Linux). Most of us use grep just for finding the words
in a file. The power of grep comes with using its options and regularexpressions. You can analyze large
sets of log files with the help of grep command.

Grep stands for Global search for Regular Expressions and Print.

The basic syntax of grep command is

grep [options] pattern [list of files]

Let see some practical examples on grep command.

1. Running the last executed grep command

This saves a lot of time if you are executing the same command again and again.

!grep

This displays the last executed grep command and also prints the result set of the command on the
terminal.

2. Search for a string in a file

This is the basic usage of grep command. It searches for the given string in the specified file.

grep "Error" logfile.txt

This searches for the string "Error" in the log file and prints all the lines that has the word "Error".

3. Searching for a string in multiple files.

grep "string" file1 file2

grep "string" file_pattern

This is also the basic usage of the grep command. You can manually specify the list of files you want to
search or you can specify a file pattern (use regular expressions) tosearch for.

4. Case insensitive search

The -i option enables to search for a string case insensitively in the give file. It matches the words like
"UNIX", "Unix", "unix".

grep -i "UNix" file.txt

5. Specifying the search string as a regular expression pattern.

grep "^[0-9].*" file.txt

This will search for the lines which starts with a number. Regular expressions is huge topic and I am not
covering it here. This example is just for providing the usage of regularexpressions.

6. Checking for the whole words in a file.

By default, grep matches the given string/pattern even if it found as a substring in a file. The -w option
to grep makes it match only the whole words.

grep -w "world" file.txt

7. Displaying the lines before the match.

Some times, if you are searching for an error in a log file; it is always good to know the lines around the
error lines to know the cause of the error.

grep -B 2 "Error" file.txt

This will prints the matched lines along with the two lines before the matched lines.

8. Displaying the lines after the match.

grep -A 3 "Error" file.txt

This will display the matched lines along with the three lines after the matched lines.

9. Displaying the lines around the match

grep -C 5 "Error" file.txt

This will display the matched lines and also five lines before and after the matched lines.

10. Searching for a sting in all files recursively

You can search for a string in all the files under the current directory and sub-directories with the help -r
option.

grep -r "string" *

11. Inverting the pattern match

You can display the lines that are not matched with the specified search sting patternusing the -v option.

grep -v "string" file.txt

12. Displaying the non-empty lines

You can remove the blank lines using the grep command.

grep -v "^$" file.txt

13. Displaying the count of number of matches.

We can find the number of lines that matches the given string/pattern

grep -c "sting" file.txt

14. Display the file names that matches the pattern.

We can just display the files that contains the given string/pattern.

grep -l "string" file.txt

15. Display the file names that do not contain the pattern.

We can display the files which do not contain the matched string/pattern.

grep -l "string" file.txt

16. Displaying only the matched pattern.

By default, grep displays the entire line which has the matched string. We can make the grep to display
only the matched string by using the -o option.

grep -o "string" file.txt

17. Displaying the line numbers.

We can make the grep command to display the position of the line which contains the matched string in
a file using the -n option

grep -n "string" file.txt

18. Displaying the position of the matched string in the line

The -b option allows the grep command to display the character position of the matched string in a file.

grep -o -b "string" file.txt

19. Matching the lines that start with a string

The ^ regular expression pattern specifies the start of a line. This can be used in grep to match the lines
which start with the given string or pattern.

grep "^start" file.txt

20. Matching the lines that end with a string

The $ regular expression pattern specifies the end of a line. This can be used in grep to match the lines
which end with the given string or pattern.

grep "end$" file.txt

Basic Unix and Linux Commands With Examples
Learning unix operating system is very easy. It is just that you need to understand the unix server
concepts and familiar with the unix commands. Here I am providing some important unix
commands which will be used in daily work.

Unix Commands With Examples:

1. Listing files

The first thing after logging into the unix system, everyone does is listing the files in a directory. The ls
command is used to list the files in a directory.

>ls

add.sh

logfile.txt

prime.pl

If you simply execute ls on the command prompt, then it will display the files and directories in the
current directory.

>ls /usr/local/bin

You can pass a directory as an argument to ls command. In this case, the ls command prints all the files
and directories in the specific directory you have passed.

2. Displaying the contents of a file.

The next thing is to display the contents of a file. The cat command is used to display thecontents in a
file.

>cat file.txt

This is a sample unix file

Learning about unix server is awesome

3. Displaying first few lines from a file.

The head command can be used to print the specified number of lines from the starting of a file. The
below head command displays the first five lines of file.

>head -5 logfile.dat

4. Displaying last few lines from a file.

The tail command can be used to print the specified number of lines from the ending of a file. The
below tail command displays the last three lines of file.

>tail -3 logfile.dat

5. Changing the directories

The cd command can be used to change from one directory to another directory. You need to specify
the target directory where you want to go.

>cd /var/tmp

After typing this cd command you will be in /var/tmp directory.

6. Creating a file.

The touch command simply creates an empty file. The below touch command creates a new file in the
current directory.

touch new_file.txt

7. copying the contents of one file into another.

The cp command is used to copy the content of source file into the target file. If the target file already
have data, then it will be overwritten.

>cp source_file target_file

8. Creating a directory.

Directories are a way of organizing your files. The mkdir command is used to create the specified
directory.

>mkdir backup

This will create the backup directory in the current directory.

9. Renaming and moving the files.

The mv command is used to rename the files and it also used for moving the files from one directory
into another directory.

Renaming the file.

>mv file.txt new_file.txt

Moving the file to another directory.

>mv new_file.txt tmp/

10. Finding the number of lines in a file

The wc command can be used to find the number of line, words and characters in a file.

>wc logfile.txt

21 26 198 logfile.txt

To know about the unix command, it is always good to see the man pages. To see the man pages
simply pass the command as an argument to the man.

man ls

SQL Queries Interview Questions - Oracle Analytical Functions Part 1
Analytic functions compute aggregate values based on a group of rows. They differ from aggregate
functions in that they return multiple rows for each group. Most of the SQL developers won't use
analytical functions because of its cryptic syntax or uncertainty about its logic of operation. Analytical
functions saves lot of time in writing queries and gives better performance when compared to native
SQL.

Before starting with the interview questions, we will see the difference between the aggregate functions
and analytic functions with an example. I have used SALES TABLEas an example to solve
the interview questions. Please create the below sales table in your oracle database.

CREATE TABLE SALES

(

SALE_ID INTEGER,

PRODUCT_ID INTEGER,

YEAR INTEGER,

Quantity INTEGER,

PRICE INTEGER

);










INSERT INTO SALES VALUES ( 10,200, 2009, 14, 9000);






COMMIT;

SELECT * FROM SALES;

SALE_ID PRODUCT_ID YEAR QUANTITY PRICE

--------------------------------------

1 100 2008 10 5000

2 100 2009 12 5000

3 100 2010 25 5000

4 100 2011 16 5000

5 100 2012 8 5000

6 200 2010 10 9000

7 200 2011 15 9000

8 200 2012 20 9000

9 200 2008 13 9000

10 200 2009 14 9000

11 300 2010 20 7000

12 300 2011 18 7000

13 300 2012 20 7000

14 300 2008 17 7000

15 300 2009 19 7000

Difference Between Aggregate and Analytic Functions:

Q. Write a query to find the number of products sold in each year?

The SQL query Using Aggregate functions is

SELECT Year,

COUNT(1) CNT

FROM SALES

GROUP BY YEAR;

YEAR CNT

---------

2009 3

2010 3

2011 3

2008 3

2012 3

The SQL query Using Aanalytic functions is

SELECT SALE_ID,

PRODUCT_ID,

Year,

QUANTITY,

PRICE,

COUNT(1) OVER (PARTITION BY YEAR) CNT

FROM SALES;

SALE_ID PRODUCT_ID YEAR QUANTITY PRICE CNT

------------------------------------------

9 200 2008 13 9000 3

1 100 2008 10 5000 3

14 300 2008 17 7000 3

15 300 2009 19 7000 3

2 100 2009 12 5000 3

10 200 2009 14 9000 3

11 300 2010 20 7000 3

6 200 2010 10 9000 3

3 100 2010 25 5000 3

12 300 2011 18 7000 3

4 100 2011 16 5000 3

7 200 2011 15 9000 3

13 300 2012 20 7000 3

5 100 2012 8 5000 3

8 200 2012 20 9000 3

From the ouputs, you can observe that the aggregate functions return only one row per group whereas
analytic functions keeps all the rows in the gorup. Using the aggregate functions, the select clause
contains only the columns specified in group by clause and aggregate functions whereas in analytic
functions you can specify all the columns in thetable.

The PARTITION BY clause is similar to GROUP By clause, it specifies the window of rows that the
analytic funciton should operate on.

I hope you got some basic idea about aggregate and analytic functions. Now lets start with solving
the Interview Questions on Oracle Analytic Functions.

1. Write a SQL query using the analytic function to find the total sales(QUANTITY) of each product?

Solution:

SUM analytic function can be used to find the total sales. The SQL query is

SELECT PRODUCT_ID,

QUANTITY,

SUM(QUANTITY) OVER( PARTITION BY PRODUCT_ID ) TOT_SALES

FROM SALES;

PRODUCT_ID QUANTITY TOT_SALES

-----------------------------

100 12 71

100 10 71

100 25 71

100 16 71

100 8 71

200 15 72

200 10 72

200 20 72

200 14 72

200 13 72

300 20 94

300 18 94

300 17 94

300 20 94

300 19 94

2. Write a SQL query to find the cumulative sum of sales(QUANTITY) of each product? Here first sort
the QUANTITY in ascendaing order for each product and then accumulate the QUANTITY.
Cumulative sum of QUANTITY for a product = QUANTITY of current row + sum of QUANTITIES all
previous rows in that product.

Solution:

We have to use the option "ROWS UNBOUNDED PRECEDING" in the SUM analytic function to get the
cumulative sum. The SQL query to get the ouput is

SELECT PRODUCT_ID,

QUANTITY,

SUM(QUANTITY) OVER( PARTITION BY PRODUCT_ID

ORDER BY QUANTITY ASC

ROWS UNBOUNDED PRECEDING) CUM_SALES

FROM SALES;

PRODUCT_ID QUANTITY CUM_SALES

-----------------------------

100 8 8

100 10 18

100 12 30

100 16 46

100 25 71

200 10 10

200 13 23

200 14 37

200 15 52

200 20 72

300 17 17

300 18 35

300 19 54

300 20 74

300 20 94

The ORDER BY clause is used to sort the data. Here the ROWS UNBOUNDED PRECEDING option
specifies that the SUM analytic function should operate on the current row and the pervious rows
processed.

3. Write a SQL query to find the sum of sales of current row and previous 2 rows in a product group?
Sort the data on sales and then find the sum.

Solution:

The sql query for the required ouput is

SELECT PRODUCT_ID,

QUANTITY,

SUM(QUANTITY) OVER(

PARTITION BY PRODUCT_ID

ORDER BY QUANTITY DESC

ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) CALC_SALES

FROM SALES;

PRODUCT_ID QUANTITY CALC_SALES

------------------------------

100 25 25

100 16 41

100 12 53

100 10 38

100 8 30

200 20 20

200 15 35

200 14 49

200 13 42

200 10 37

300 20 20

300 20 40

300 19 59

300 18 57

300 17 54

The ROWS BETWEEN clause specifies the range of rows to consider for calculating the SUM.

4. Write a SQL query to find the Median of sales of a product?

Solution:

The SQL query for calculating the median is

SELECT PRODUCT_ID,

QUANTITY,

PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY QUANTITY ASC)

OVER (PARTITION BY PRODUCT_ID) MEDIAN

FROM SALES;

PRODUCT_ID QUANTITY MEDIAN

--------------------------

100 8 12

100 10 12

100 12 12

100 16 12

100 25 12

200 10 14

200 13 14

200 14 14

200 15 14

200 20 14

300 17 19

300 18 19

300 19 19

300 20 19

300 20 19

5. Write a SQL query to find the minimum sales of a product without using the group by clause.

Solution:

The SQL query is

SELECT PRODUCT_ID,

YEAR,

QUANTITY

FROM

(

SELECT PRODUCT_ID,

YEAR,

QUANTITY,

ROW_NUMBER() OVER(PARTITION BY PRODUCT_ID

ORDER BY QUANTITY ASC) MIN_SALE_RANK

FROM SALES

) WHERE MIN_SALE_RANK = 1;

PRODUCT_ID YEAR QUANTITY

------------------------

100 2012 8

200 2010 10

300 2008 17

SQL Interview Questions and Answers
1. What is Normalization?

Normalization is the process of organizing the columns, tables of a database to minimize the
redundancy of data. Normalization involves in dividing large tables into smaller tables and defining
relationships between them. Normalization is used in OLTP systems.

2. What are different types of Normalization Levels or Normalization Forms?

The different types of Normalization Forms are:

• First Normal Form: Duplicate columns from the same table needs to be eliminated. We have to
create separate tables for each group of related data and identify each row with a unique column or
set of columns (Primary Key)

• Second Normal Form: First it should meet the requirement of first normal form. Removes the
subsets of data that apply to multiple rows of a table and place them in separate tables.
Relationships must be created between the new tables and their predecessors through the use of
foreign keys.

• Third Normal Form: First it should meet the requirements of second normal form. Remove columns
that are not depending upon the primary key.

• Fourth Normal Form: There should not be any multi-valued dependencies.

Most databases will be in Third Normal Form

3. What is De-normalization?

De-normalization is the process of optimizing the read performance of a database by adding redundant
data or by grouping data. De-normalization is used in OLAP systems.

4. What is a Transaction?

A transaction is a logical unit of work performed against a database in which all steps must be
performed or none.

5. What are ACID properties?

A database transaction must be Atomic, Consistent, Isolation and Durability.

• Atomic: Transactions must be atomic. Transactions must fail or succeed as a single unit.

• Consistent: The database must always be in consistent state. There should not be any partial
transactions

• Isolation: The changes made by a user should be visible only to that user until the transaction is
committed.

• Durability: Once a transaction is committed, it should be permanent and cannot be undone.

6. Explain different storage models of OLAP?

• MOLAP: The data is stored in multi-dimensional cube. The storage is not in the relational database,
but in proprietary formats.

• ROLAP: ROLAP relies on manipulating the data stored in the RDBMS for slicing and dicing
functionality.

• HOLAP: HOLAP combines the advantages of both MOLAP and ROLAP. For summary type
information, HOLAP leverages on cube technology for faster performance. For detail information,
HOLAP can drill through the cube.

7. Explain one-to-one relationship with an example?

One to one relationship is a simple reference between two tables. Consider Customer and Address
tables as an example. A customer can have only one address and an address references only one
customer.

8. Explain one-to-many relationship with an example?

One-to-many relationships can be implemented by splitting the data into two tables with a primary key
and foreign key relationship. Here the row in one table is referenced by one or more rows in the other
table. An example is the Employees and Departments table, where the row in the Departments table is
referenced by one or more rows in the Employees table.

9. Explain many-to-many relationship with an example?

Many-to-Many relationship is created between two tables by creating a junction table with the key from
both the tables forming the composite primary key of the junction table.

An example is Students, Subjects and Stud_Sub_junc tables. A student can opt for one or more
subjects in a year. Similarly a subject can be opted by one or more students. So a junction table is
created to implement the many-to-many relationship.

10. Write down the general syntax of a select statement?

The basic syntax of a select statement is

SELECT Columns | *

FROM Table_Name

[WHERE Search_Condition]

[GROUP BY Group_By_Expression]

[HAVING Search_Condition]

[ORDER BY Order_By_Expression [ASC|DESC]]

String aggregating Analytic Functions in Oracle Database
The string aggregate functions concatenate multiple rows into a single row. Consider the
products table as an example.

Table Name: Products

Year product

-------------

2010 A

2010 B

2010 C

2010 D

2011 X

2011 Y

2011 Z

Here, in the output we will concatenate the products in each year by a comma separator. The desired
output is:

year product_list

------------------

2010 A,B,C,D

2011 X,Y,Z

LISTAGG analytic function in 11gR2:

The LISTAGG function can be used to aggregate the strings. You can pass the explicit delimiter to the
LISTAGG function.

SELECT year,

LISTAGG(product, ',') WITHIN GROUP (ORDER BY product) AS product_list

FROM products

GROUP BY year;

WM_CONCAT function:

You cannot pass an explicit delimiter to the WM_CONCAT function. It uses comma as the string
separator.

SELECT year,

wm_concat(product) AS product_list

FROM products

GROUP BY year;

Pivot and Unpivot Operators in Oracle Database 11g
Pivot:

The pviot operator converts row data to column data and also can do aggregates whileconverting. To
see how pivot operator works, consider the following "sales" table as anyexample

Table Name: Sales

customer_id product price

--------------------------------------

1 A 10

1 B 20

2 A 30

2 B 40

2 C 50

3 A 60

3 B 70

3 C 80

The rows of the "sales" table needs to be converted into columns as shown below

Table Name: sales_rev

cutomer_id a_product b_product c_product

-----------------------------------------

1 10 20

2 30 40 50

3 60 70 80

The query for converting the rows to columns is

SELECT *

FROM (SELECT customer_id,product,price from sales)

pivot ( sum(price) as total_price for (product) IN ( 'A' as a, 'B' as b, 'C' as c) )

Pivot can be used to generate the data in xml format. The query for generating the data into xml fomat
is shown below.

SELECT *

FROM (SELECT customer_id,product,price from sales)

pivot XML ( sum(price) as total_price for (product) IN ( SELECT distinct product from sales) )

If you are not using oracle 11g database, then you can implement the unpivot feature asconverting rows
to columns

Unpivot:

Unpivot operator converts the columns into rows.
Table Name: sales_rev

cutomer_id a_product b_product c_product

-----------------------------------------

1 10 20

2 30 40 50

3 60 70 80

Table Name: sales

customer_id product price

---------------------------

1 A 10

1 B 20

2 A 30

2 B 40

2 C 50

3 A 60

3 B 70

3 C 80

The query to convert rows into columns is

SELECT *

FROM sales_rev

UNPIVOT [EXCLUDE NULLs | INCLUDE NULLs] (price FOR product IN (a_product AS 'A', b_product
AS 'B', c_product_c AS 'C'));

Points to note about the query

• The columns price and product in the unpivot clause are required and these names need not to be
present in the table.

• The unpivoted columns must be specified in the IN clause

• By default the query excludes null values.

Min and Max values of contiguous rows - Oracle SQL Query
Q) How to find the Minimum and maximum values of continuous sequence numbers in a group of rows.

I know the problem is not clear without giving an example. Let say I have the Employees table with the
below data.

Table Name: Employees

Dept_Id Emp_Seq

---------------

10 1

10 2

10 3

10 5

10 6

10 8

10 9

10 11

20 1

20 2

I want to find the minimum and maximum values of continuous Emp_Seq numbers. The output should
look as.

Dept_Id Min_Seq Max_Seq

-----------------------

10 1 3

10 5 6

10 8 9

10 11 11

20 1 2

Write an SQL query in oracle to find the minimum and maximum values of continuous Emp_Seq in
each department?

STEP1: First we will generate unique sequence numbers in each department using the Row_Number
analytic function in the Oracle. The SQL query is.

SELECT Dept_Id,

Emp_Seq,

ROW_NUMBER() OVER (PARTITION BY Dept_Id ORDER BY Emp_Seq) rn

FROM employees;

Dept_Id Emp_Seq rn

--------------------

10 1 1

10 2 2

10 3 3

10 5 4

10 6 5

10 8 6

10 9 7

10 11 8

20 1 1

20 2 2

STEP2: Subtract the value of rn from emp_seq to identify the continuous sequences as a group. The
SQL query is

SELECT Dept_Id,

Emp_Seq,

Emp_Seq-ROW_NUMBER() OVER (PARTITION BY Dept_Id ORDER BY Emp_Seq) Dept_Split

FROM employees;

Dept_Id Emp_Seq Dept_Split

---------------------------

10 1 0

10 2 0

10 3 0

10 5 1

10 6 1

10 8 2

10 9 2

10 11 3

20 1 0

20 2 0

STEP3: The combination of the Dept_Id and Dept_Split fields will become the group for continuous
rows. Now use group by on these fields and find the min and max values. The final SQL query is

SELECT Dept_Id,

MIN(Emp_Seq) Min_Seq,

MAX(Emp_Seq) Max_Seq

FROM

(

SELECT Dept_Id,

Emp_Seq,

Emp_Seq-ROW_NUMBER() OVER (PARTITION BY Dept_Id ORDER BY Emp_Seq) Dept_Split

FROM employees;

)A

Group BY Dept_Id, Dept_Split

Rewrite Sql Query | Sql Performance Tuning
Tuning an SQL query for performance is a big topic. Here I will just cover how to re-write a query and
thereby improve the performance. Rewriting an SQL query is one of the ways you can improve
performance. You can rewrite a query in many different ways.

To explain this, i have used the sales and products table.

SALES(SALE_ID, YEAR, PRODUCT_ID, PRICE);
PRODUCTS(PRODUCT_ID, PRODUCT_NAME);

Follow the below steps in re writing a query for optimization.

1. Avoid Redundant Logic

I have seen people writing redundant sub-queries and worrying about their query performance. As an
example, find the total sales in each year and also the sales of product with id 10 in each year.

SELECT T.YEAR,

T.TOT_SAL,

P.PROD_10_SAL

(

SELECT YEAR,

SUM(PRICE) TOT_SAL

FROM SALES

GROUP BY YEAR

)T

LEFT OUTER JOIN

(

SELECT YEAR,

SUM(PRICE) PROD_10_SAL

FROM SALES

WHERE PRODUCT_ID = 10

)P

ON (T.YEAR = P.YEAR);

Most SQL developers write the above Sql query without even thinking that it can be solved in a single
query. The above query is rewritten as

SELECT YEAR,

SUM(CASE WHEN PRODUCT_ID = 10

THEN PRICE

ELSE NULL

END ) PROD_10_SAL,

SUM(SALES) TOT_SAL

FROM SALES

GROUP BY YEAR;

Now you can see the difference, just by reading the sales table one time we will able to solve
the problem.

First take a look at of your query, identify the redundant logic and then tune it.

2. LEFT OUTER JOIN, NOT EXISTS, NOT IN

Some times you can rewrite a LEFT OUTER JOIN by using NOT EXISTS or NOT IN and vice versa. As
an example, I want to find the products which do not sold in the year 2011.

SELECT P.PRODUCT_ID,

P.PRODUCT_NAME

FROM PRODUCTS P

LEFT OUTER JOIN

SALES S

ON (P.PRODUCT_ID = S.PRODUCT_ID)

WHERE S.SALE_ID IS NULL;

The same query can be rewritten using NOT EXISTS and NOT IN as


P.PRODUCT_NAME

FROM PRODUCTS P

WHERE NOT EXISTS

(

SELECT 1

FROM SALES S

WHERE S.PRODUCT_ID = P.PRODUCT_ID);


P.PRODUCT_NAME

FROM PRODUCTS P

WHERE PRODUCT_ID NOT IN

(

SELECT PRODUCT_ID

FROM SALES

);

Analyze the performance of these three queries and use the appropriate one.

Note: Be careful while using the NOT IN. If the sub query returns at lease row with NULL data, then the
main query won't return a row at all.

3. INNER JOIN, EXISTS, IN

As similar to LEFT OUTER JOIN, the INNER JOINS can also be implemented with the EXISTS or
IN operators. As an example, we will find the sales of products whose product id’s exists in the
products table.

SELECT S.PRODUCT_ID,

SUM(PRICE)

FROM SALES S

JOIN

PRODUCTS P

ON (S.PRODUCT_ID = P.PRODUCT_ID)

GROUP BY S.PRODUCT_ID;

As we are not selecting any columns from the products table, we can rewrite the same query with the
help of EXISTS or IN operator.


SUM(PRICE)

FROM SALES S

WHERE EXISTS

(

SELECT 1

FROM PRODUCTS P

WHERE P.PRODUCT_ID = S.PRODUCT_ID);



SUM(PRICE)

FROM SALES S

WHERE PRODUCT_ID IN

(

SELECT PRODUCT_ID

FROM PRODUCTS P

);


4. INNER JOIN, CORRELATED QUERY

We will see a simple join between the SALES and PRODUCTS table.

SELECT S.SALE_ID,

S.PRODUCT_ID,

P.PRODUCT_NAME

FROM SALES S

JOIN

PRODUCTS P

ON (S.PRODUCT_ID = P.PRODUCT_ID)

The above query can be rewritten with correlated query as

SELECT S.SALE_ID,

S.PRODUCT_ID,

(SELECT PRODUCT_NAME

FROM PRODUCTS P

WHERE P.PRODUCT_ID = S.PRODUCT_ID)

FROM SALES S

Analyze these two queries with the explain plan and check which one gives better performance.

5. Using With Clause or Temporary Tables.

Try to avoid writing complex Sql queries. Split the queries and store the data in temporary tables or use
the Oracle With Clause for temporary storage. This will improve the performance. You can also use the
temporary tables or with clause when you want to reuse the same query more than once. This saves
the time and increases the performance.

Tips for increasing the query performance:

• Create the required indexes. In the mean time avoid creating too many indexes on a table.

• Rewrite the Sql query.

• Use the explain plan, auto trace to know about the query execution.

• Generate statistics on tables.

• Specify the oracle Hints in the query.

• Ask the DBA to watch the query and gather stats like CPU usage, number of row read etc.

Top Unix Interview Questions - Part 1
1. How to display the 10th line of a file?
head -10 filename | tail -1

2. How to remove the header from a file?
sed -i '1 d' filename

3. How to remove the footer from a file?
sed -i '$ d' filename

4. Write a command to find the length of a line in a file?
The below command can be used to get a line from a file.
sed –n '<n> p' filename
We will see how to find the length of 10th line in a file
sed -n '10 p' filename|wc -c

5. How to get the nth word of a line in Unix?
cut –f<n> -d' '

6. How to reverse a string in unix?
echo "java" | rev

7. How to get the last word from a line in Unix file?
echo "unix is good" | rev | cut -f1 -d' ' | rev

8. How to replace the n-th line in a file with a new line in Unix?
sed -i'' '10 d' filename # d stands for delete
sed -i'' '10 i new inserted line' filename # i stands for insert

9. How to check if the last command was successful in Unix?
echo $?

10. Write command to list all the links from a directory?
ls -lrt | grep "^l"

11. How will you find which operating system your system is running on in UNIX?
uname -a

12. Create a read-only file in your home directory?
touch file; chmod 400 file

13. How do you see command line history in UNIX?
The 'history' command can be used to get the list of commands that we are executed.

14. How to display the first 20 lines of a file?
By default, the head command displays the first 10 lines from a file. If we change the option of head,
then we can display as many lines as we want.
head -20 filename
An alternative solution is using the sed command
sed '21,$ d' filename
The d option here deletes the lines from 21 to the end of the file

15. Write a command to print the last line of a file?

The tail command can be used to display the last lines from a file.
tail -1 filename
Alternative solutions are:
sed -n '$ p' filename
awk 'END{print $0}' filename

1. How do you rename the files in a directory with _new as suffix?
ls -lrt|grep '^-'| awk '{print "mv "$9" "$9".new"}' | sh

2. Write a command to convert a string from lower case to upper case?
echo "apple" | tr [a-z] [A-Z]

3. Write a command to convert a string to Initcap.
echo apple | awk '{print toupper(substr($1,1,1)) tolower(substr($1,2))}'

4. Write a command to redirect the output of date command to multiple files?
The tee command writes the output to multiple files and also displays the output on the terminal.
date | tee -a file1 file2 file3

5. How do you list the hidden files in current directory?
ls -a | grep '^.'

6. List out some of the Hot Keys available in bash shell?

• Ctrl+l - Clears the Screen.

• Ctrl+r - Does a search in previously given commands in shell.

• Ctrl+u - Clears the typing before the hotkey.

• Ctrl+a - Places cursor at the beginning of the command at shell.

• Ctrl+e - Places cursor at the end of the command at shell.

• Ctrl+d - Kills the shell.

• Ctrl+z - Places the currently running process into background.

7. How do you make an existing file empty?
cat /dev/null > filename

8. How do you remove the first number on 10th line in file?
sed '10 s/[0-9][0-9]*//' < filename

9. What is the difference between join -v and join -a?
join -v : outputs only matched lines between two files.
join -a : In addition to the matched lines, this will output unmatched lines also.

10. How do you display from the 5th character to the end of the line from a file?
cut -c 5- filename

1. Display all the files in current directory sorted by size?
ls -l | grep '^-' | awk '{print $5,$9}' |sort -n|awk '{print $2}'

2. Write a command to search for the file 'map' in the current directory?
find -name map -type f

3. How to display the first 10 characters from each line of a file?
cut -c -10 filename

4. Write a command to remove the first number on all lines that start with "@"?
sed ',^@, s/[0-9][0-9]*//' < filename

5. How to print the file names in a directory that has the word "term"?
grep -l term *
The '-l' option make the grep command to print only the filename without printing the content of the file.
As soon as the grep command finds the pattern in a file, it prints the pattern and stops searching other
lines in the file.

6. How to run awk command specified in a file?
awk -f filename

7. How do you display the calendar for the month march in the year 1985?
The cal command can be used to display the current month calendar. You can pass the month and year
as arguments to display the required year, month combination calendar.
cal 03 1985
This will display the calendar for the March month and year 1985.

8. Write a command to find the total number of lines in a file?
wc -l filename
Other ways to pring the total number of lines are
awk 'BEGIN {sum=0} {sum=sum+1} END {print sum}' filename
awk 'END{print NR}' filename

9. How to duplicate empty lines in a file?
sed '/^$/ p' < filename

10. Explain iostat, vmstat and netstat?

• Iostat: reports on terminal, disk and tape I/O activity.

• Vmstat: reports on virtual memory statistics for processes, disk, tape and CPU activity.

• Netstat: reports on the contents of network data structures.

cat file1 file2 file3 > file

2. How to display the fields in a text file in reverse order?
awk 'BEGIN {ORS=""} { for(i=NF;i>0;i--) print $i," "; print "n"}' filename

3. Write a command to find the sum of bytes (size of file) of all files in a directory.
ls -l | grep '^-'| awk 'BEGIN {sum=0} {sum = sum + $5} END {print sum}'

4. Write a command to print the lines which end with the word "end"?
grep 'end$' filename
The '$' symbol specifies the grep command to search for the pattern at the end of the line.

5. Write a command to select only those lines containing "july" as a whole word?
grep -w july filename
The '-w' option makes the grep command to search for exact whole words. If the specified pattern is
found in a string, then it is not considered as a whole word. For example: In the string "mikejulymak",
the pattern "july" is found. However "july" is not a whole word in that string.

6. How to remove the first 10 lines from a file?
sed '1,10 d' < filename

7. Write a command to duplicate each line in a file?
sed 'p' < filename

8. How to extract the username from 'who am i' comamnd?
who am i | cut -f1 -d' '

9. Write a command to list the files in '/usr' directory that start with 'ch' and then display the number of
lines in each file?
wc -l /usr/ch*
Another way is
find /usr -name 'ch*' -type f -exec wc -l {} ;

10. How to remove blank lines in a file ?
grep -v ‘^$’ filename > new_filename

ps -aef | grep <user_name>

2. Write a command to display all the files recursively with path under current directory?
find . -depth -print

3. Display zero byte size files in the current directory?
find -size 0 -type f

4. Write a command to display the third and fifth character from each line of a file?
cut -c 3,5 filename

5. Write a command to print the fields from 10th to the end of the line. The fields in the line are delimited
by a comma?
cut -d',' -f10- filename

6. How to replace the word "Gun" with "Pen" in the first 100 lines of a file?
sed '1,00 s/Gun/Pen/' < filename

7. Write a Unix command to display the lines in a file that do not contain the word "RAM"?
grep -v RAM filename
The '-v' option tells the grep to print the lines that do not contain the specified pattern.

8. How to print the squares of numbers from 1 to 10 using awk command
awk 'BEGIN { for(i=1;i<=10;i++) {print "square of",i,"is",i*i;}}'

9. Write a command to display the files in the directory by file size?
ls -l | grep '^-' |sort -nr -k 5

10. How to find out the usage of the CPU by the processes?
The top utility can be used to display the CPU usage by the processes.

The basename utility deletes any prefix ending in /. The usage is mentioned below:
basename /usr/local/bin/file
This will display only file

2. How to display zero byte size files?
ls -l | grep '^-' | awk '/^-/ {if ($5 !=0 ) print $9 }'

3. How to replace the second occurrence of the word "bat" with "ball" in a file?
sed 's/bat/ball/2' < filename

4. How to remove all the occurrences of the word "jhon" except the first one in a line with in the entire
file?
sed 's/jhon//2g' < filename

5. How to replace the word "lite" with "light" from 100th line to last line in a file?
sed '100,$ s/lite/light/' < filename

6. How to list the files that are accessed 5 days ago in the current directory?
find -atime 5 -type f

7. How to list the files that were modified 5 days ago in the current directory?
find -mtime 5 -type f

8. How to list the files whose status is changed 5 days ago in the current directory?
find -ctime 5 -type f

9. How to replace the character '/' with ',' in a file?
sed 's///,/' < filename
sed 's|/|,|' < filename

10. Write a command to find the number of files in a directory.
ls -l|grep '^-'|wc -l

The Yes utility can be used to repeatedly output a line with the specified string or 'y'.
yes <your_name> | head -100

2. Write a command to display the first 10 characters from each line of a file?
cut -c -10 filename

3. The fields in each line are delimited by comma. Write a command to display third field from each line
of a file?
cut -d',' -f2 filename

4. Write a command to print the fields from 10 to 20 from each line of a file?
cut -d',' -f10-20 filename

5. Write a command to print the first 5 fields from each line?
cut -d',' -f-5 filename

6. By default the cut command displays the entire line if there is no delimiter in it. Which cut option is
used to supress these kind of lines?
The -s option is used to supress the lines that do not contain the delimiter.

7. Write a command to replace the word "bad" with "good" in file?
sed s/bad/good/ < filename

8. Write a command to replace the word "bad" with "good" globally in a file?
sed s/bad/good/g < filename

9. Write a command to replace the word "apple" with "(apple)" in a file?
sed s/apple/(&)/ < filename

10. Write a command to switch the two consecutive words "apple" and "mango" in a file?

sed 's/(apple) (mango)/2 1/' < filename

11. Write a command to display the characters from 10 to 20 from each line of a file?
cut -c 10-20 filename

1. Write a command to print the lines that has the the pattern "july" in all the files in a particular
directory?
grep july *
This will print all the lines in all files that contain the word “july” along with the file name. If any of the
files contain words like "JULY" or "July", the above command would not print those lines.

2. Write a command to print the lines that has the word "july" in all the files in a directory and also
suppress the filename in the output.
grep -h july *

3. Write a command to print the lines that has the word "july" while ignoring the case.
grep -i july *
The option i make the grep command to treat the pattern as case insensitive.

4. When you use a single file as input to the grep command to search for a pattern, it won't print the
filename in the output. Now write a grep command to print the filename in the output without using the '-
H' option.
grep pattern filename /dev/null
The /dev/null or null device is special file that discards the data written to it. So, the /dev/null is always
an empty file.
Another way to print the filename is using the '-H' option. The grep command for this is
grep -H pattern filename

5. Write a command to print the file names in a directory that does not contain the word "july"?
grep -L july *
The '-L' option makes the grep command to print the filenames that do not contain the specified pattern.

6. Write a command to print the line numbers along with the line that has the word "july"?
grep -n july filename
The '-n' option is used to print the line numbers in a file. The line numbers start from 1

7. Write a command to print the lines that starts with the word "start"?
grep '^start' filename
The '^' symbol specifies the grep command to search for the pattern at the start of the line.

8. In the text file, some lines are delimited by colon and some are delimited by space. Write a command
to print the third field of each line.
awk '{ if( $0 ~ /:/ ) { FS=":"; } else { FS =" "; } print $3 }' filename

9. Write a command to print the line number before each line?
awk '{print NR, $0}' filename

10. Write a command to print the second and third line of a file without using NR.
awk 'BEGIN {RS="";FS="n"} {print $2,$3}' filename

11. How to create an alias for the complex command and remove the alias?
The alias utility is used to create the alias for a command. The below command createsalias for ps -aef
command.
alias pg='ps -aef'
If you use pg, it will work the same way as ps -aef.
To remove the alias simply use the unalias command as
unalias pg

12. Write a command to display todays date in the format of 'yyyy-mm-dd'?
The date command can be used to display todays date with time
date '+%Y-%m-%d'

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