Why filtering behavior matters in PostgreSQL - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all rows where the age is greater than 30.

PostgreSQL

SELECT * FROM users WHERE age [1] 30;

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

A<

B=

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' will select ages less than 30, which is wrong.

Using '=' will select only age exactly 30, not greater.

The '>' operator filters rows where age is greater than 30.

2fill in blank

medium

Complete the code to filter users whose name starts with 'A'.

PostgreSQL

SELECT * FROM users WHERE name [1] 'A%';

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

ALIKE

B=

CILIKE

DIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' will look for exact match 'A%', not pattern.

Using IN is for lists, not patterns.

The LIKE operator filters strings matching a pattern. 'A%' means names starting with 'A'.

3fill in blank

hard

Fix the error in the query to filter users with age not equal to 25.

PostgreSQL

SELECT * FROM users WHERE age [1] 25;

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

A<>

B!=

C==

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' causes syntax error.

Using '=' selects equal values, not 'not equal'.

In PostgreSQL, '<>' means 'not equal'. '!=' also works but '<>' is standard SQL.

4fill in blank

hard

Fill both blanks to select users with age between 20 and 30 inclusive.

PostgreSQL

SELECT * FROM users WHERE age [1] 20 AND age [2] 30;

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

A>=

B<

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' or '<' excludes boundary ages 20 or 30.

Mixing operators incorrectly causes wrong filtering.

Use '>=' for age at least 20 and '<=' for age at most 30 to include both ends.

5fill in blank

hard

Fill all three blanks to select users whose name starts with 'J', age is greater than 25, and city is not 'London'.

PostgreSQL

SELECT * FROM users WHERE name [1] 'J%' AND age [2] 25 AND city [3] 'London';

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

ALIKE

B>

C!=

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of '!=' for city filters wrong rows.

Using '=' instead of LIKE for name pattern causes no matches.

Use LIKE for pattern matching name starting with 'J', '>' for age greater than 25, and '!=' for city not equal to 'London'.