June 28, 2017

Oracle SQL Complex Queries Frequently Asked In Interview Questions


Oracle SQL Complex Queries Frequently Asked In various Interviews

SELECT * FROM PRODUCTS;


Products Table:

PRODUCT_ID    PRODUCT_NAME

100                            Nokia
200                            IPhone
300                            Samsung
400                            LG


Sales Table:

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


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.

STEP 2: 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
(
SELECT P.PRODUCT_NAME,
       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.

Method 1: 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

Method 2: Using the NOT IN operator.

SELECT P.PRODUCT_NAME
FROM   PRODUCTS P
WHERE  P.PRODUCT_ID NOT IN
       (SELECT DISTINCT PRODUCT_ID FROM SALES);

PRODUCT_NAME
------------
LG

Method 3: Using the NOT EXISTS operator.

SELECT P.PRODUCT_NAME
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 with the sales in the year 2011. The SQL query to get the required output is

SELECT P.PRODUCT_NAME
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
(
SELECT P.PRODUCT_NAME,
       S.YEAR,
       RANK() OVER (
              PARTITION BY S.YEAR
              ORDER BY S.QUANTITY DESC
              ) RNK
FROM   PRODUCTS P,
       SALES S
WHERE  P.PRODUCT_ID = S.PRODUCT_ID
) 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.

SELECT P.PRODUCT_NAME,
       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

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

Solution:

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

SELECT P.PRODUCT_NAME,
       S.YEAR,
       S.QUANTITY
FROM   PRODUCTS P,
       SALES S
WHERE  P.PRODUCT_ID = S.PRODUCT_ID
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

7. 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

8. 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

SELECT P.PRODUCT_NAME,
       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

9. 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
(
SELECT P.PRODUCT_NAME,
       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.

SELECT P.PRODUCT_NAME,
       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;

10. 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

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

Solution:

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

12. Write a query to display only Sunday 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') = 'SUN';

13. 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;

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

B
I
G

Solution:

SELECT SUBSTR('SMILE',LEVEL,1) A
FROM   DUAL
CONNECT BY LEVEL <=LENGTH('BIG');

15. 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.

Solution 1:

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')
);


16. 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;

17. 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 output 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);

18. 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;

19. This is an extension to the problem 18. 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;

20. This is an extension to the problem 18. write a query to get the top 5 products in each year 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;

21. 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 should not 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;

22. 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 table structure. 

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)
);

23. How to retrieve 2nd highest sal in each departement from emp and dept tables using GROUP BY?

SELECT e.DeptNo, MAX(e.Sal),d.DeptName Salary


FROM Emp e left outer join dept d ON e.DeptNo=d.DeptNo


WHERE e.Sal <


(SELECT MAX(Sal)


 FROM Emp


 WHERE DeptNo = e.DeptNo)


GROUP BY e.DeptNo,d.DeptName

24. Find the 3rd MAX and MIN salary in the emp table?

select distinct sal from emp e1 where 3 = (select count(distinct sal) from emp e2 where e1.sal <= e2.sal);

select distinct sal from emp e1 where 3 = (select count(distinct sal) from emp e2where e1.sal >= e2.sal);

25. If there are two tables emp1 and emp2, and both have common record. How can I fetch all the records but common records only once?

(Select * from emp) Union (Select * from emp1)

How to fetch only common records from two tables emp and emp1?

(Select * from emp) Intersect (Select * from emp1)

How can I retrive all records of emp1 those should not present in emp2?

(Select * from emp) Minus (Select * from emp1)



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