WHERE with LIKE pattern matching in SQL - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select all customers whose names start with 'A'.

SQL

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

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

ALIKE

B=

CIN

DBETWEEN

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of LIKE causes exact match only.

Using IN or BETWEEN does not support pattern matching.

The LIKE operator is used to match patterns in SQL. Here, 'A%' means names starting with 'A'.

2fill in blank

medium

Complete the code to find products with 'book' anywhere in their description.

SQL

SELECT * FROM products WHERE description [1] '%book%';

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

A=

BBETWEEN

CIN

DLIKE

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' only matches exact descriptions.

Using IN or BETWEEN does not work for pattern matching.

The LIKE operator with '%book%' finds any description containing 'book' anywhere.

3fill in blank

hard

Fix the error in the query to find emails ending with '.com'.

SQL

SELECT email FROM users WHERE email [1] '%.com';

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

A=

BBETWEEN

CLIKE

DIN

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' causes no matches unless exact string is '.com'.

Using IN or BETWEEN is invalid for pattern matching.

The LIKE operator is needed to match the pattern '%.com' for emails ending with '.com'.

4fill in blank

hard

Fill both blanks to select orders where the status starts with 'sh' and ends with 'ed'.

SQL

SELECT * FROM orders WHERE status [1] 'sh%' AND status [2] '%ed';

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

ALIKE

B=

CIN

DBETWEEN

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' causes exact match only, so no partial matches.

Using IN or BETWEEN is not valid for pattern matching.

Both conditions use LIKE with wildcards to match patterns at the start and end of the status.

5fill in blank

hard

Fill all three blanks to select employees whose first name contains 'an', last name starts with 'S', and email ends with '.org'.

SQL

SELECT * FROM employees WHERE first_name [1] '%an%' AND last_name [2] 'S%' AND email [3] '%.org';

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

ALIKE

B=

CIN

DBETWEEN

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' causes exact matches only.

Using IN or BETWEEN is invalid for pattern matching.

All three conditions use LIKE with appropriate patterns to match parts of the strings.