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.

Practice

1. Which SQL clause is used to filter rows based on a pattern in a text column?
SELECT * FROM employees WHERE name LIKE 'A%';

easy

A. GROUP BY

B. ORDER BY

C. WHERE with LIKE

D. JOIN

Solution

1. Step 1: Understand the purpose of LIKE

   The LIKE keyword is used to match text patterns in SQL queries.

2. Step 2: Identify the filtering clause

   The WHERE clause filters rows, and combined with LIKE, it filters rows matching a pattern.

3. Final Answer:

   WHERE with LIKE -> Option C

4. Quick Check:

   WHERE with LIKE = B [OK]

Hint: LIKE is always used with WHERE to filter text patterns [OK]

Common Mistakes:

• Confusing LIKE with GROUP BY
• Using LIKE without WHERE
• Thinking ORDER BY filters rows

2. Which of the following is the correct syntax to find names starting with 'Jo' in a table called users?

easy

A. SELECT * FROM users WHERE name LIKE '%Jo';

B. SELECT * FROM users WHERE name LIKE 'Jo%';

C. SELECT * FROM users WHERE name LIKE '_Jo%';

D. SELECT * FROM users WHERE name LIKE 'Jo_';

Solution

1. Step 1: Understand the pattern for names starting with 'Jo'

   Names starting with 'Jo' means 'Jo' followed by any characters, so pattern is 'Jo%'.

2. Step 2: Match the correct syntax

   SELECT * FROM users WHERE name LIKE 'Jo%'; uses LIKE 'Jo%' which matches names starting with 'Jo'. Others do not match this pattern.

3. Final Answer:

   SELECT * FROM users WHERE name LIKE 'Jo%'; -> Option B

4. Quick Check:

   Start with 'Jo' = LIKE 'Jo%' [OK]

Hint: Use % after text to match any ending characters [OK]

Common Mistakes:

• Using % before text to match start
• Using _ instead of %
• Misplacing wildcards

3. Given the table products with a column product_code, what rows will this query return?

SELECT * FROM products WHERE product_code LIKE '_A%';

medium

A. Rows where product_code starts with 'A'

B. Rows where product_code ends with 'A'

C. Rows where product_code contains 'A' anywhere

D. Rows where product_code has 'A' as the second character

Solution

1. Step 1: Understand the pattern '_A%'

   The underscore (_) matches exactly one character, so the first character can be anything. 'A' is the second character. '%' means any characters after.

2. Step 2: Interpret the query result

   The query returns rows where the second character of product_code is 'A', regardless of the first character or what follows.

3. Final Answer:

   Rows where product_code has 'A' as the second character -> Option D

4. Quick Check:

   _A% means second char is 'A' [OK]

Hint: Use _ for single char, % for any chars [OK]

Common Mistakes:

• Thinking _ matches zero or multiple characters
• Confusing position of 'A' in pattern
• Assuming LIKE matches anywhere without wildcards

4. Identify the error in this SQL query:

SELECT * FROM customers WHERE city LIKE '%York_';

It should find cities ending with 'York' followed by exactly one character.

medium

A. The pattern is correct; no error

B. The underscore should be before 'York' like '_%York'

C. The pattern should be '%York%' to match any after 'York'

D. The underscore should be after 'York' but without % like '%York_'

Solution

1. Step 1: Analyze the pattern '%York_'

   The % matches any characters before 'York', then 'York' literal, then _ matches exactly one character after.

2. Step 2: Check if it matches cities ending with 'York' plus one char

   This pattern correctly matches cities ending with 'York' plus exactly one character, so the query is correct.

3. Final Answer:

   The pattern is correct; no error -> Option A

4. Quick Check:

   '%York_' matches 'York' + 1 char [OK]

Hint: Check if _ and % are placed as intended [OK]

Common Mistakes:

• Thinking _ matches multiple characters
• Confusing placement of % and _
• Assuming pattern must end with %

5. You want to find all employees whose email ends with '.com' but have exactly one character before '.com'. Which query is correct?

hard

A. SELECT * FROM employees WHERE email LIKE '%_.com';

B. SELECT * FROM employees WHERE email LIKE '%_.com%';

C. SELECT * FROM employees WHERE email LIKE '%.__.com';

D. SELECT * FROM employees WHERE email LIKE '%.com';

Solution

1. Step 1: Understand the requirement

   The email should end with '.com' and have exactly one character before '.com'.

2. Step 2: Analyze each pattern

   SELECT * FROM employees WHERE email LIKE '%.com'; matches any email ending with '.com' but no restriction on characters before.
   SELECT * FROM employees WHERE email LIKE '%_.com'; uses '%_.com' meaning any characters, then exactly one character, then '.com' at the end.
   SELECT * FROM employees WHERE email LIKE '%.__.com'; expects two characters before '.com' which is incorrect.
   SELECT * FROM employees WHERE email LIKE '%_.com%'; allows any characters after '.com' which is incorrect for ending.

3. Final Answer:

   SELECT * FROM employees WHERE email LIKE '%_.com'; -> Option A

4. Quick Check:

   Use _ before '.com' for exactly one char [OK]

Hint: Use _ before '.com' to match exactly one char [OK]

Common Mistakes:

• Using % instead of _ for single character
• Placing % after '.com' allowing extra chars
• Miscounting characters before '.com'