SQL Interview Questions: 1. Write a query to find the highest
salary earned by an employee in each department and also the number
of employees who earn the highest salary?
SELECT DEPARTMENT_ID, MAX(SALARY) HIGHEST_SALARY, COUNT(1)
KEEP(DENSE_RANK LAST ORDER BY SALARY) CNT_HIGH_SAL FROM EMPLOYEES
GROUP BY DEPARTMENT_ID;
2. Write a query to get the top 2 employees who are earning the
highest salary in each department?
SELECT DEPARTMENT_ID, EMPLOYEE_ID, SALARY FROM ( SELECT
DEPARTMENT_ID, EMPLOYEE_ID, SALARY, ROW_NUMBER() OVER(PARTITION BY
DEPARTMENT_ID ORDER BY SALARY DESC ) R FROM EMPLOYEES ) A WHERE R (
SELECT AVG(SALARY) FROM EMPLOYEES E_I WHERE E_I.DEPARTMENT_ID =
E_O.DEPARTMENT_ID );
5. How do you display the current date in oracle?
SELECT SYSDATE FROM DUAL;
6. What is a correlated Query? It is a form of sub query, where
the sub query uses the values from the outer query in its WHERE
clause. The sub query runs for each row processed in the outer
query. Question 4 is an example for a correlated sub query.
PL/SQL Interview Questions: 1. What is a cursor? A cursor is a
reference to the system memory when an SQL statement is executed. A
cursor contains the information about the select statement and the
rows accessed by it. 2. What is implicit cursor and explicit
cursor?
Implicit Cursors: Implicit cursors are created by default when
DML statements like INSERT, UPDATE and DELETE are executed in
PL/SQL objects. Explicit Cursors: Explicit cursors must be created
by you when executing the select statements.
3. What are the attributes of a cursor?
Cursor attributes are:
%FOUND : Returns true if a DML or SELECT statement affects at
least one row. %NOTFOUND: Returns true if a DML or SELECT statement
does not affect at least one row. %ROWCOUNT: Returns the number of
rows affected by the DML or SELECT statement. %ISOPEN: Returns true
if a cursor is in open state. %BULK_ROWCOUNT: Similar to %ROWCOUNT,
except it is used in bulk operations.
4. What is a private and public procedure?
Public procedure: In a package, the signature of the procedure
is specified in the package specification. This procedure can be
called outside of the package. Private procedure: For private
procedure, there wont be any signature in the package
specification. So, these procedures can be called only inside the
package and cannot be called outside of the package.
5. Create a sample delete trigger on employees table?
CREATE OR REPLACE TRIGGER EMPLOYEES_AD" AFTER DELETE ON
EMPLOYEES
REFERENCING NEW AS NEW OLD AS OLD FOR EACH ROW
BEGIN INSERT INTO employees_changes (employee_id, change_date )
VALUES (:OLD.photo_tag_id, SYSDATE ); END;
SQL Interview Questions and Answers1. 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 isSELECT Columns | * FROM
[WHERE Table_Name Search_Condition]
[GROUP BY Group_By_Expression] [HAVING Search_Condition] [ORDER
BY Order_By_Expression [ASC|DESC]]
1. Load the below products table into the target table.CREATE
TABLE PRODUCTS ( PRODUCT_ID PRODUCT_NAME ); INTEGER,
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 ----------------------100 200 300 400
500 600 Nokia IPhone Samsung LG BlackBerry 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
isCREATE TABLE TGT_PRODUCTS ( PRODUCT_ID PRODUCT_NAME INSERT_DATE
); INTEGER, VARCHAR2(30), 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
WHERE ) ORDER BY DBMS_RANDOM.VALUE --Random number generator in
oracle. )A WHERE ROWNUM (SELECT AVG(QUANTITY) FROM SALES S1 WHERE
S1.PRODUCT_ID = S.PRODUCT_ID );
PRODUCT_NAME YEAR QUANTITY -------------------------Nokia IPhone
Samsung Samsung 2010 2012 2012 2010 25 20 20 20
2. Write a query to compare the products sales of "IPhone" and
"Samsung" in each year? The output should look like asYEAR
IPHONE_QUANT SAM_QUANT IPHONE_PRICE SAM_PRICE
---------------------------------------------------
2010 2011 2012
10 15 20
20 18 20
9000 9000 9000
7000 7000 7000
Solution: By using self-join SQL query we can get the required
result. The required SQL query isSELECT S_I.YEAR, S_I.QUANTITY
IPHONE_QUANT, S_S.QUANTITY SAM_QUANT, S_I.PRICE S_S.PRICE FROM
PRODUCTS P_I, SALES S_I, PRODUCTS P_S, SALES S_S WHERE AND AND AND
AND P_I.PRODUCT_ID = S_I.PRODUCT_ID P_S.PRODUCT_ID = S_S.PRODUCT_ID
P_I.PRODUCT_NAME = 'IPhone' P_S.PRODUCT_NAME = 'Samsung' S_I.YEAR =
S_S.YEAR IPHONE_PRICE, SAM_PRICE
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 IPhone IPhone Nokia Nokia
Nokia Samsung Samsung Samsung 2011 2012 2010 2012 2011 2010 2010
2012 2011 0.333333333 0.444444444 0.222222222 0.163265306
0.326530612 0.510204082 0.344827586 0.344827586 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
asPRODUCT_NAME QUAN_2010 QUAN_2011 QUAN_2012
-----------------------------------------IPhone 10 15 20
Samsung Nokia
20 25
18 16
20 8
Solution: Oracle 11g provides a pivot function to transpose the
row data into column data. The SQL query for this isSELECT * 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 isSELECT YEAR,
COUNT(1) NUM_PRODUCTS FROM SALES
GROUP BY YEAR;
YEAR
NUM_PRODUCTS
-----------------2010 2011 2012 3 3 3
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 PRODUCT_NAME ); INTEGER, VARCHAR2(30)
CREATE TABLE SALES ( SALE_ID PRODUCT_ID YEAR Quantity PRICE );
INTEGER, INTEGER, INTEGER, INTEGER, 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 ----------------------100 200 300 Nokia
IPhone Samsung
The sales table contains the following data.SELECT * FROM
SALES;
SALE_ID PRODUCT_ID YEAR QUANTITY PRICE
-------------------------------------1 2 3 4 5 6 7 8 9 100 100 100
200 200 200 300 300 300 2010 2011 2012 2010 2011 2012 2010 2011
2012 25 16 8 10 15 20 20 18 20 5000 5000 5000 9000 9000 9000 7000
7000 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 isSELECT 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 Nokia Nokia IPhone
2012 2011 2010 2012 8 16 25 20 16 25 0 15
IPhone IPhone Samsung Samsung Samsung
2011 2010 2012 2011 2010
15 10 20 18 20
10 0 18 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 years
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 isSELECT 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 )A GROUP BY PRODUCT_NAME HAVING
MIN(QUAN_DIFF) >= 0; P.PRODUCT_ID = S.PRODUCT_ID
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 WHERE (P.PRODUCT_ID = S.PRODUCT_ID); S.QUANTITY IS
NULL
PRODUCT_NAME -----------LG
Method2: Using the NOT IN operator.SELECT P.PRODUCT_NAME FROM
WHERE PRODUCTS P 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 WHERE PRODUCTS P 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 isSELECT
P.PRODUCT_NAME FROM PRODUCTS P, SALES S_2012, SALES S_2011 WHERE
AND P.PRODUCT_ID = S_2012.PRODUCT_ID S_2012.YEAR = 2012
AND AND AND
S_2011.YEAR = 2011 S_2012.PRODUCT_ID = S_2011.PRODUCT_ID
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 isSELECT 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 Samsung IPhone
Samsung 2010 2011 2012 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 Samsung Nokia
405000 406000 245000
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 = a.l order by t.id, a.l;
How to find (calculate) median using oracle sql queryA 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
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 ouput 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 The source data is represented in the form the tree
structure. You can easily derive the parent-child relationship
between the elements. For example, B is parent of D and E. As the
element A is root element, it is at level 0. B, C are at level 1
and so on.
The above tree structure data is represented in a table as shown
below. c1, A, A, A, A, A, c2, B, B, B, C, C, c3, D, D, E, F, G, c4
H I NULL NULL NULL
Here in this table, column C1 is parent of column C2, column C2
is parent of column C3, column C3 is parent of column C4. Q1. Write
a query to load the target table with the below data. Here you need
to generate sequence numbers for each element and then you have to
get the parent id. As the element "A" is at root, it does not have
any parent and its parent_id is NULL. id, element, lev, parent_id
1, A, 0, NULL 2, B, 1, 1 3, C, 1, 1 4, D, 2, 2 5, E, 2, 2 6, F, 2,
3 7, G, 2, 3 8, H, 3, 4 9, I, 3, 4 Solution: WITH t1 AS ( SELECT
VALUE PARENT, LEV, LEAD(value,1) OVER (PARTITION BY r ORDER BY lev)
CHILD FROM (SELECT c1, c2, c3, c4,
ROWNUM r FROM table_name ) UNPIVOT (value FOR lev IN (c1 as 0,c2
as 1,c3 as 2,c4 as 3)) ), t2 AS ( SELECT PARENT, LEV, ROWNUM SEQ
FROM (SELECT DISTINCT PARENT, LEV FROM T1 ORDER BY LEV ) ), T3 AS (
SELECT DISTINCT PARENT, CHILD FROM T1 WHERE CHILD IS NOT NULL UNION
ALL SELECT DISTINCT NULL, PARENT FROM T1 WHERE LEV=0 ) SELECT C.SEQ
Id, T3.CHILD ELEMENT, C.LEV, P.SEQ PARENT_ID FROM T3 INNER JOIN T2
C ON (T3.CHILD = C.PARENT) LEFT OUTER JOIN T2 P ON (T3.PARENT =
P.PARENT) ORDER BY C.SEQ; 1. Consider the following friends table
as the sourceName, Friend_Name ----------------sam, sam, ram
vamsi
vamsi, ram
vamsi, jhon ram, ram, vijay 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 asName, Friend_of_Firend
---------------------sam, sam, sam, sam, ram jhon vijay anand
Solution:SELECT f1.name, f2.friend_name as friend_of_friend FROM
friends f1, friends f2 WHERE AND f1.name = 'sam' 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 asName,
Friend_of_Friend
---------------------sam, sam, sam, jhon vijay anand
Solution:SELECT f1.name, f2.friend_name as friend_of_friend FROM
friends f1, friends f2 WHERE AND AND f1.name = 'sam' f1.friend_name
= f2.name 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 asProducts, quantity_sold, year
----------------------------A, B, C, D, 200, 155, 455, 620, 2009
2009 2009 2009
E, F, G, H, I, J, L, M,
135, 390, 999, 810, 910, 109, 260, 580,
2009 2009 2010 2010 2010 2010 2010 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
