Oracle SQL Queries Most Frequently Asked In Check Point Software Written Test Interview
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');
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')
);
Solution:
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');
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')
);
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;
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);
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;
 |
| Oracle SQL Queries Most Frequently Asked In Check Point Software Written Test Interview |
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;
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;
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;
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)
);
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
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);
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)
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
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);
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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