Scalar subqueries in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the name and the highest salary from the employees table using a scalar subquery.

MySQL

SELECT name, (SELECT [1] FROM employees) AS max_salary FROM employees;

Drag options to blanks, or click blank then click option'

ACOUNT(salary)

BMAX(salary)

Csalary

DMIN(salary)

Attempts:

3 left

💡 Hint

Common Mistakes

Using salary without aggregation returns multiple rows causing an error.

Using COUNT(salary) returns the number of salaries, not the highest salary.

The scalar subquery must return the maximum salary, so MAX(salary) is correct.

2fill in blank

medium

Complete the code to find employees whose salary is equal to the average salary using a scalar subquery.

MySQL

SELECT name FROM employees WHERE salary = (SELECT [1] FROM employees);

Drag options to blanks, or click blank then click option'

AAVG(salary)

BMIN(salary)

CSUM(salary)

DMAX(salary)

Attempts:

3 left

💡 Hint

Common Mistakes

Using MAX(salary) or MIN(salary) returns wrong comparison values.

Using SUM(salary) returns the total salary, not average.

The scalar subquery should return the average salary, so AVG(salary) is correct.

3fill in blank

hard

Fix the error in the scalar subquery to correctly find employees with salary greater than the average salary.

MySQL

SELECT name FROM employees WHERE salary > (SELECT [1]salary FROM employees);

Drag options to blanks, or click blank then click option'

AMAX(

BFROM

CAVG(

DMIN(

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting AVG() parentheses causes syntax errors.

Using FROM without an aggregate function is invalid here.

The scalar subquery must use AVG(salary) to get the average salary for comparison.

4fill in blank

hard

Fill both blanks to select employees whose salary is between the minimum and maximum salary using scalar subqueries.

MySQL

SELECT name FROM employees WHERE salary BETWEEN (SELECT [1](salary) FROM employees) AND (SELECT [2](salary) FROM employees);

Drag options to blanks, or click blank then click option'

AMIN

BMAX

CAVG

DCOUNT

Attempts:

3 left

💡 Hint

Common Mistakes

Using AVG() or COUNT() returns incorrect ranges.

Swapping MIN and MAX changes the logic.

The salary should be between the minimum and maximum salaries, so MIN and MAX are correct.

5fill in blank

hard

Fill all three blanks to select employees whose salary is greater than the average salary but less than the maximum salary.

MySQL

SELECT name FROM employees WHERE salary > (SELECT [1](salary) FROM employees) AND salary < (SELECT [2](salary) FROM employees) ORDER BY [3];

Drag options to blanks, or click blank then click option'

AMIN

BAVG

Csalary

DMAX

Attempts:

3 left

💡 Hint

Common Mistakes

Using MIN() instead of AVG() or MAX() causes wrong filtering.

Ordering by an aggregate function instead of a column causes errors.

The query compares salary to average and maximum salaries, so AVG and MAX are correct. Ordering by salary sorts the results by salary.