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 examples
on 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.
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);
INSERT INTO SALES VALUES ( 2, 100, 2011, 16, 5000);
INSERT INTO SALES VALUES ( 3, 100, 2012, 8, 5000);
INSERT INTO SALES VALUES ( 4, 200, 2010, 10, 9000);
INSERT INTO SALES VALUES ( 5, 200, 2011, 15, 9000);
INSERT INTO SALES VALUES ( 6, 200, 2012, 20, 9000);
INSERT INTO SALES VALUES ( 7, 300, 2010, 20, 7000);
INSERT INTO SALES VALUES ( 8, 300, 2011, 18, 7000);
INSERT INTO SALES VALUES ( 9, 300, 2012, 20, 7000);
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 below SQL 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
(
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.
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.
SELECT P.PRODUCT_NAME
FROM PRODUCTS P
WHERE P.PRODUCT_ID NOT IN
(SELECT DISTINCT
PRODUCT_ID FROM SALES);
PRODUCT_NAME
------------
LG
Method3: 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
1. 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
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
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
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
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
(
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;
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
1. Write a query to
generate sequence numbers from 1 to the specified number N?
Solution:
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 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;
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 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
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 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)
);
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 TABLE as an example to solve the interview questions. Please create the below sales table in your oracle database.
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 TABLE as 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 ( 1, 100, 2008, 10, 5000);
INSERT INTO SALES VALUES ( 2, 100, 2009, 12, 5000);
INSERT INTO SALES VALUES ( 3, 100, 2010, 25, 5000);
INSERT INTO SALES VALUES ( 4, 100, 2011, 16, 5000);
INSERT INTO SALES VALUES ( 5, 100, 2012, 8, 5000);
INSERT INTO SALES VALUES ( 6, 200, 2010, 10, 9000);
INSERT INTO SALES VALUES ( 7, 200, 2011, 15, 9000);
INSERT INTO SALES VALUES ( 8, 200, 2012, 20, 9000);
INSERT INTO SALES VALUES ( 9, 200, 2008, 13, 9000);
INSERT INTO SALES VALUES ( 10,200, 2009, 14, 9000);
INSERT INTO SALES VALUES ( 11, 300, 2010, 20, 7000);
INSERT INTO SALES VALUES ( 12, 300, 2011, 18, 7000);
INSERT INTO SALES VALUES ( 13, 300, 2012, 20, 7000);
INSERT INTO SALES VALUES ( 14, 300, 2008, 17, 7000);
INSERT INTO SALES VALUES ( 15, 300, 2009, 19, 7000);
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 the table.
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
Oracle
Query to split the delimited data in a column to multiple rows
1.
Consider the following table "t" data as the source
id value
----------
1 A,B,C
2 P,Q,R,S,T
3 M,N
Here the data in value column is a delimited by comma. Now write a query to split the delimited data in the value column into multiple rows. The output should look like as
id value
--------
1 A
1 B
1 C
2 P
2 Q
2 R
2 S
2 T
3 M
3 N
Solution:
SELECT t.id,
CASE WHEN a.l = 1
THEN substr(value, 1, instr(value,',',1,a.l)-1)
ELSE substr(value, instr(value,',',1,a.l-1)+1,
CASE WHEN instr(value,',',1,a.l)-instr(value,',',1,a.l-1)-1 > 0
THEN instr(value,',',1,a.l)-instr(value,',',1,a.l-1)-1
ELSE length(value)
END
)
END final_value
FROM t,
( SELECT level l
FROM DUAL
CONNECT BY LEVEL <=
(
SELECT Max(length(value) - length(replace(value,',',''))+1) FROM t
)
) a
WHERE length(value) - length(replace(value,',',''))+1 >= a.l
order by t.id, a.l;
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.
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
How to
find (calculate) median using oracle sql query
A median
is a value separating the higher half of sample from the lower half. The median
can be found by arranging all the numerical values from lowest to highest value
and picking the middle one. If there are even number of numerical values, then
there is no single middle value; then the median is defined as the mean of the
two middle values.
Now let see how to calculate the median in oracle with the employees table as example.
Table name: employees
Now let see how to calculate the median in oracle with the employees table as example.
Table name: employees
empid, deptid, salary
1, 100, 5000
2, 100, 3000
3, 100, 4000
5, 200, 6000
6, 200, 8000
The below query is used to calculate the median of employee salaries across the entire table.
select empid,
dept_id,
salary,
percentile_disc(0.5) within group (order by salary desc)
over () median
from employees;
The output of the above query is
empid, deptid, salary, median
-----------------------------
1, 100, 5000, 5000
2, 100, 3000, 5000
3, 100, 4000, 5000
5, 200, 6000, 5000
6, 200, 8000, 5000
Now we
will write a query to find the median of employee salaries in each department.
select empid,
dept_id,
salary,
percentile_disc(0.5) within group (order by salary desc)
over (partition by department_id) median
from employees;
The output of the above query is
empid, deptid, salary, median
------------------------------
1, 100, 5000, 4000
2, 100, 3000, 4000
3, 100, 4000, 4000
5, 200, 6000, 7000
6, 200, 8000, 7000
Oracle
Query to Repeat a Number n Times
I need to
write one query which will accept input parameter from user and display that
number of times the user input. For example if I give input parameter as 5 then
it should display 55555, if I give 4 it should give output as 4444.
This question was asked by one of my blog reader. Here is the oracle sql query.
Solution
This question was asked by one of my blog reader. Here is the oracle sql query.
Solution
SELECT sys_connect_by_path('','5')
FROM dual
WHERE level = 5
CONNECT BY level <= 5
Here the connect by level clause is used to repeat the rows n number of times. In our case it repeats the row 5 times. As dual table contains only one record, the connect by clause repeats that single row 5 times.
SYS_CONNECT_BY_PATH is valid only in hierarchical queries. It returns the path of a column value from root to node, with column values separated by char (In this example empty string) for each row returned by CONNECT BY condition. In our case, it concatenates all the 5's in each row.
The condition level=5 is used to get only the last row.
Different
Ways (How) to Delete Duplicate Rows in Table
In this
article I am going to show different ways of deleting duplicating records from
the table. It is a common question in interviews which is asked frequently.
Consider the following table with rows as an example:
Consider the following table with rows as an example:
Table Name: Products
ProductId Price
---------------
1 10
1 10
2 20
3 30
3 30
Here assume that productId column should be unique after deleting. Now we see how to delete the duplicate records from the products table in different ways.
1. Using Rowid
The following Delete statement deletes the rows using the Rowid.
Syntax:
Delete from <tablename>
where rowid not in (select max(rowid) from <tablename> group by <unique columns or primary key>);
Example:
Delete from products
where rowid not in (select max(rowid) from products group by productid);
2. Using temp table and Distinct
Here, first create a temp table and insert distinct rows in the temp table. Then truncate the main table and insert records from the temp table.
Create temporary table products_temp As
Select Distinct ProductID, Price
From Products;
Truncate table Products;
Insert into products
Select *
From products_temp;
3. Using temp table and Row Number analytic function.
The row_number analytic function is used to rank the rows. Here we use the row_number function to rank the rows for each group of productId and then select only record from the group.
Create temporary table products_temp As
Select productid, price
From
(
Select productid, price,
row_number() over (partition by productId order by price) group_rank
From products
)
Where group_rank = 1;
Please comment here if you know any other methods of deleting the duplicate records from the table.
SQL
Query to Select Numeric Values from Varchar Column in Oracle
Problem
Description: One of the columns in the table is of varchar data type. It
contains both strings, numbers and mix of char and numeric values. My problem
is to print only the values that contains only numeric digits.
Sample data in the columne is shown below:
Sample data in the columne is shown below:
varchar_column
--------------
Ora123
786
database
92db
The output should contain the following values:
786
Solution1: Using regexp_like function
The following oracle sql query uses regexp_like function to get only the values that are purely numeric:
select varchar_column
from table_name
where regexp_like(varchar_column,'^[0-9]$');
Solution2: Using translate function
Another way is using both the translate and length function. The sql query is shown below:
SELECT varchar_column
FROM table_name
WHERE length(translate(varchar_column,'0123456789','1')) is null
AND varchar_column IS NOT NULL;
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
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 while converting. To see how pivot operator works, consider the following "sales" table as any example
The pviot operator converts row data to column data and also can do aggregates while converting. To see how pivot operator works, consider the following "sales" table as any example
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.
If you like this post,
then please share it by clicking on the +1 button.
Oracle
Query to split the delimited data in a column to multiple rows
1.
Consider the following table "t" data as the source
id value
----------
1 A,B,C
2 P,Q,R,S,T
3 M,N
Here the data in value column is a delimited by comma. Now write a query to split the delimited data in the value column into multiple rows. The output should look like as
id value
--------
1 A
1 B
1 C
2 P
2 Q
2 R
2 S
2 T
3 M
3 N
Solution:
SELECT t.id,
CASE WHEN a.l = 1
THEN substr(value, 1, instr(value,',',1,a.l)-1)
ELSE substr(value, instr(value,',',1,a.l-1)+1,
CASE WHEN instr(value,',',1,a.l)-instr(value,',',1,a.l-1)-1 > 0
THEN instr(value,',',1,a.l)-instr(value,',',1,a.l-1)-1
ELSE length(value)
END
)
END final_value
FROM t,
( SELECT level l
FROM DUAL
CONNECT BY LEVEL <=
(
SELECT Max(length(value) - length(replace(value,',',''))+1) FROM t
)
) a
WHERE length(value) - length(replace(value,',',''))+1 >= a.l
order by t.id, a.l;
Convert
String to Ascii Values in Oracle
The below
query converts a string to ascii characters.
select replace(substr(dump('oracle'),instr(dump('oracle'),': ')+2),',') from dual;
The output of this query is 1111149799108101.
select replace(substr(dump('oracle'),instr(dump('oracle'),': ')+2),',') from dual;
The output of this query is 1111149799108101.
Converting
Rows to Columns
Lets see the conversion of rows to columns with an example.
Suppose we have a products table which looks like
Table: products
Table: products
product_id
|
product_name
|
1
|
AAA
|
1
|
BBB
|
1
|
CCC
|
2
|
PPP
|
2
|
QQQ
|
2
|
RRR
|
Now we want to convert the data in the products table as
product_id
|
prodcut_name_1
|
prodcut_name_2
|
prodcut_name_3
|
1
|
AAA
|
BBB
|
CCC
|
2
|
PPP
|
QQQ
|
RRR
|
The following query converts the rows to columns:
SELECT product_id,
MAX(DECODE(product_id,1,product_name,NULL))
product_name_1,
MAX(DECODE(product_id,1,product_name,NULL))
product_name_2,
MAX(DECODE(product_id,1,product_name,NULL))
product_name_3
FROM products
GROUP BY product_id;
Query
to Generate Sequence numbers 1 to n
In oracle
we can generate sequence numbers from 1 to n by using the below query:
SELECT rownum
FROM dual
CONNECT BY LEVEL<=n;
Replace n with a number.
SELECT rownum
FROM dual
CONNECT BY LEVEL<=n;
Replace n with a number.
Converting
Columns to Rows in Oracle
Lets see
the conversion of columns to rows with an example. Suppose we have a table
which contains the subjects handled by each teacher. The table looks like
Table: teachers
Table: teachers
teacher_id |
subject1 |
subject2 |
subject3 |
1 |
maths |
physics |
english |
2 |
social |
science |
drawing |
Now we want to convert the data in the teachers table as
teacher_id
|
subject
|
1
|
maths
|
1
|
physics
|
1
|
english
|
2
|
social
|
2
|
science
|
2
|
drawing
|
To achieve this we need each row in teachers table to be repeated 3 times (number of subject columns). The following query converts the columns into rows:
SELECT teacher_id,
CASE pivot
WHEN 1
THEN subject1
WHEN 2
THEN subject2
WHEN 3
THEN subject3
ELSE NULL
END subject
FROM teachers,
(SELECT rownum pivot from dual
CONNECT BY LEVEL <=3)
Concatenating
multiple rows into a single column dynamically - Oracle
Q) How to
concatenate multiple rows of a column in a table into a single column?
I have to concatenate multiple rows to a single column. For example consider the below teachers table.
I have to concatenate multiple rows to a single column. For example consider the below teachers table.
Table Name: Teacher
Teacher_id subject_name
-----------------------
1 Biology
1 Maths
1 Physics
2 English
2 Social
The above table is a normalized table containing the subjects and teacher id. We will denormalize the table, by concatenating the subjects of each teacher into a single column and thus preserving the teacher id as unique in the output. The output data should look like as below
teacher_id subjects_list
-------------------------------
1 Biology|Maths|Physics
2 English|Social
How to achieve this?
Solution:
We can concatenate multiple rows in to a single column dynamically by using the Hierarchical query. The SQL query to get the result is
SELECT teacher_id,
SUBSTR(SYS_CONNECT_BY_PATH(subject_name, '|'),2)
subjects_list
FROM
(
SELECT teacher_id,
subject_name,
COUNT(*) OVER (PARTITION BY teacher_id) sub_cnt,
ROW_NUMBER () OVER (PARTITION BY teacher_id
ORDER BY subject_name) sub_seq
FROM teachers
) A
WHERE sub_seq=sub_cnt
START WITH sub_seq=1
CONNECT BY prior sub_seq+1=sub_seq
AND prior teacher_id=teacher_id
No comments:
Post a Comment