HOME BLOG PORTFOLIO PHOTO CONTACT
mysql tricky query

1. Consider a table named “salary” having the following columns:

“id” (type: INT)
“salary” (type: INT)
“incentive” (type: INT)
“tax” (type: INT)

Write a standard SQL query which will update the tax column with the sum of 10% of salary and 2% of incentive, for those salaries which are more than 15000.

UPDATE salary SET tax = 0.1*salary+0.02*incentive WHERE salary > 15000

2. Consider a table named “employee” having the following columns:

“empid” (type: INT)
“empname” (type: TEXT)
“salary” (type: INT)

Write a standard SQL query which retrieves the empnames whose values start with the string ‘john’ followed by any characters.

SELECT empname FROM employee WHERE empname LIKE 'john%'

3. Consider a table named “employee” having the following columns:

“empid” (type: INT)
“empname” (type: TEXT)
“salary” (type: INT)

Write a standard SQL query which retrieves the number of rows where the salary is not null. The returned value should be represented using the column name “validsalarycount”.
1     SELECT count(salary) AS validsalarycount FROM employee WHERE salary IS NOT NULL

4. Consider a table named “store” having the following columns:

“storename” (type: TEXT)
“sales” (type: INT)
“Date” (type: DATE)

Write a standard SQL query which retrieves the storenames, whose sales lie between 100 and 2000 (not inclusive). The storenames should not be repeated.

SELECT DISTINCT storename FROM  store WHERE sales > 100 AND sales < 2000

5. Consider a table named “staff” having the following column structure:

“empid” (type: INT)
“empname” (type: TEXT)
“salary” (type: INT)

Write a standard SQL query which retrieves the sum of 75 percent of the salaries from the staff table (only salaries above 5000 are to be considered). The returned value should be represented using the column name ‘total’.

SELECT sum(0.75*salary) AS total FROM staff WHERE salary > 5000

6.Consider a database with a table called “accounts”, having two fields:

“entrydate” (type: DATE)
“accountno” (type: INT)

Write a SQL query which returns the accountno of the most recent entrydate. The returned value should be represented using the column name, “accountno”.

Correct solution:

SELECT accountno FROM accounts ORDER BY entrydate DESC LIMIT 1

!WRONG SQL!:

SELECT accountno FROM accounts ORDER BY entrydate DESC LIMIT 1

7.Consider a table called “students”, having the following column fields:

“id” (type: INT)
“name” (type: TEXT)
“marks” (type: INT)

Write a SQL query which will calculate the average of the marks of the students passing. The passing criteria is that the marks should be at least 40. The average marks are to be returned using the column name ‘marksaverage’.

SELECT avg(marks) AS marksaverage FROM students WHERE id IN (SELECT id FROM students WHERE marks >= 40)

8.Consider a table called “department”, having the following columns:

“id” (type: INT)
“deptname” (type: TEXT)
“rank” (type: INT)

Write a SQL query which will return the deptnames of the departments whose rank lies between 2 and 5 (inclusive). The results should be returned in increasing order of rank (rank 3 being higher than rank 6).

SELECT deptname FROM department WHERE rank >= 2 AND rank <= 5 ORDER BY rank ASC

9.Consider the following tables:

department
———-
deptid (type: INT)
deptname (type: TEXT)
hours (type: INT)
active (type: BIT)

employee
——–
empid (type: INT)
empname (type: TEXT)
deptid (type: INT)
designation (type: TEXT)
salary (type: INT)

Write a query to return the columns empname and deptname of the employees belonging to those departments that have a head count of 4 or more. The records should be returned in alphabetical order of empname.

This is one of the questions I did not cope with. Any solution would be appeciated.

select e.empname , d.deptname, d.`deptid`from employee e inner join department d where e.`deptid` = d.`deptid` and e.`deptid` in (select deptid from employee group by deptid having (count(`deptid`)) >= 4) order by e.empname

But this query is not optimized because i have used sub-query.

   Share on Facebook

Page views:176