COUNT function in MySQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to count all rows in the employees table.

MySQL

SELECT [1](*) FROM employees;

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

ATOTAL

BSUM

CCOUNT

DNUMBER

Attempts:

3 left

💡 Hint

Common Mistakes

Using SUM instead of COUNT.

Using TOTAL or NUMBER which are not valid SQL functions.

The COUNT function counts the number of rows. Using COUNT(*) counts all rows in the table.

2fill in blank

medium

Complete the code to count only the non-null values in the salary column.

MySQL

SELECT COUNT([1]) FROM employees;

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

Asalary

BDISTINCT salary

CNULL

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using COUNT(*) which counts all rows regardless of nulls.

Using COUNT(NULL) which always returns zero.

COUNT(column) counts only non-null values in that column. Here, COUNT(salary) counts all employees with a salary value.

3fill in blank

hard

Fix the error in the code to count distinct departments.

MySQL

SELECT COUNT([1] department) FROM employees;

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

ASINGLE

BALL

CUNIQUE

DDISTINCT

Attempts:

3 left

💡 Hint

Common Mistakes

Using ALL which counts all values including duplicates.

Using UNIQUE or SINGLE which are not valid SQL keywords here.

COUNT(DISTINCT column) counts unique non-null values in that column. DISTINCT is the correct keyword.

4fill in blank

hard

Fill both blanks to count employees with salary greater than 50000.

MySQL

SELECT COUNT([1]) FROM employees WHERE salary [2] 50000;

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

A*

Bsalary

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using COUNT(salary) which counts only non-null salaries but is less clear here.

Using < instead of > which counts salaries less than 50000.

COUNT(*) counts all rows matching the condition. The condition salary > 50000 filters employees with salary above 50000.

5fill in blank

hard

Fill all three blanks to count distinct job titles with salary less than 70000.

MySQL

SELECT COUNT([1] job_title) FROM employees WHERE salary [2] [3];

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

ADISTINCT

B<

C70000

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < which counts salaries greater than 70000.

Not using DISTINCT which counts duplicates.

COUNT(DISTINCT job_title) counts unique job titles. The condition salary < 70000 filters employees with salary less than 70000.

Practice

1. What does the COUNT(*) function do in MySQL?

easy

A. Counts all rows in the table, including those with NULL values

B. Counts only rows where all columns are NOT NULL

C. Counts only rows with NULL values in any column

D. Counts the number of columns in the table

Solution

1. Step 1: Understand COUNT(*) behavior

   The COUNT(*) function counts every row in the table regardless of NULL values.

2. Step 2: Compare with other COUNT variants

   Unlike COUNT(column), which counts only non-NULL values in that column, COUNT(*) counts all rows.

3. Final Answer:

   Counts all rows in the table, including those with NULL values -> Option A

4. Quick Check:

   COUNT(*) counts all rows [OK]

Hint: COUNT(*) counts every row, NULL or not [OK]

Common Mistakes:

• Thinking COUNT(*) ignores rows with NULL
• Confusing COUNT(*) with COUNT(column)
• Assuming COUNT(*) counts columns

2. Which of the following is the correct syntax to count non-NULL values in the column age from the table users?

easy

A. SELECT COUNT(*) FROM users;

B. SELECT COUNT(age) FROM users WHERE age != NULL;

C. SELECT COUNT(age) FROM users WHERE age = NULL;

D. SELECT COUNT(age) FROM users;

Solution

1. Step 1: Identify correct COUNT syntax for non-NULL values

   COUNT(column) counts only non-NULL values in that column, so COUNT(age) counts non-NULL ages.

2. Step 2: Check WHERE clause correctness

   SELECT COUNT(*) FROM users; uses COUNT(*) which counts all rows. Using WHERE age IS NOT NULL is valid but unnecessary with COUNT(age). Options B and C use incorrect NULL comparisons.

3. Final Answer:

   SELECT COUNT(age) FROM users; -> Option D

4. Quick Check:

   COUNT(column) counts non-NULL values [OK]

Hint: Use COUNT(column) to count non-NULL values directly [OK]

Common Mistakes:

• Using WHERE age = NULL instead of IS NOT NULL
• Using COUNT(*) to count non-NULL values
• Using != NULL which is invalid in SQL

3. Given the table orders with 5 rows where the discount column has 3 NULL values and 2 non-NULL values, what will this query return?

SELECT COUNT(discount) FROM orders;

medium

A. 5

B. 3

C. 2

D. 0

Solution

1. Step 1: Understand COUNT(column) counts non-NULL values

   The query counts only rows where discount is NOT NULL.

2. Step 2: Apply to given data

   There are 2 non-NULL discount values, so the count is 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   COUNT(discount) = 2 [OK]

Hint: COUNT(column) ignores NULLs, counts only real values [OK]

Common Mistakes:

• Counting total rows instead of non-NULL values
• Confusing NULL count with non-NULL count
• Assuming COUNT(column) counts NULLs

4. Identify the error in this query:

SELECT COUNT(name) FROM employees WHERE name = NULL;

medium

A. Using COUNT(name) instead of COUNT(*)

B. Incorrect NULL comparison with '=' instead of IS NULL

C. Missing GROUP BY clause

D. COUNT cannot be used with WHERE clause

Solution

1. Step 1: Check WHERE clause NULL comparison

   In SQL, NULL cannot be compared with '='; use IS NULL instead.

2. Step 2: Validate COUNT usage

   COUNT(name) is valid and WHERE clause is allowed; no GROUP BY needed here.

3. Final Answer:

   Incorrect NULL comparison with '=' instead of IS NULL -> Option B

4. Quick Check:

   Use IS NULL for NULL checks [OK]

Hint: Use IS NULL, not = NULL, to check for NULL values [OK]

Common Mistakes:

• Using '=' to compare NULL values
• Thinking COUNT can't be combined with WHERE
• Adding unnecessary GROUP BY

5. You have a table sales with columns region and amount. You want to find how many sales records exist for each region, including regions with zero sales. Which query will correctly show all regions with their sales count, including zero counts?

hard

A. SELECT r.region, COUNT(s.amount) FROM regions r LEFT JOIN sales s ON r.region = s.region GROUP BY r.region;

B. SELECT region, COUNT(*) FROM sales GROUP BY region;

C. SELECT region, COUNT(amount) FROM sales GROUP BY region;

D. SELECT region, COUNT(amount) FROM sales WHERE amount IS NOT NULL GROUP BY region;

Solution

1. Step 1: Understand requirement to include regions with zero sales

   To include regions with zero sales, we must use a LEFT JOIN from all regions to sales.

2. Step 2: Analyze query options

   SELECT r.region, COUNT(s.amount) FROM regions r LEFT JOIN sales s ON r.region = s.region GROUP BY r.region; uses LEFT JOIN from regions table to sales, counting sales per region including zero counts. Options A, B, D only count existing sales records, missing zero counts.

3. Final Answer:

   SELECT r.region, COUNT(s.amount) FROM regions r LEFT JOIN sales s ON r.region = s.region GROUP BY r.region; -> Option A

4. Quick Check:

   LEFT JOIN + COUNT(column) includes zero counts [OK]

Hint: Use LEFT JOIN with COUNT(column) to include zero counts [OK]

Common Mistakes:

• Using only sales table, missing regions with zero sales
• Using COUNT(*) which counts joined rows, not zero
• Not using LEFT JOIN to include all regions