WHERE with LIKE pattern matching in SQL

Introduction

The WHERE clause with LIKE helps you find rows where a text column matches a pattern. It is useful when you want to search for similar words or partial matches.

Finding all customers whose names start with 'Jo'.

Searching for products that contain the word 'book' anywhere in their name.

Filtering emails that end with '@example.com'.

Looking for addresses that include a specific street name.

Finding phone numbers that start with a certain area code.

Syntax

SQL

SELECT column1, column2 FROM table_name WHERE column_name LIKE 'pattern';

The pattern can include % which means any number of characters (including zero).

The pattern can include _ which means exactly one character.

Examples

Finds all employees whose names start with 'A'.

SQL

SELECT * FROM employees WHERE name LIKE 'A%';

Finds all products with 'book' anywhere in the description.

SQL

SELECT * FROM products WHERE description LIKE '%book%';

Finds all customers with emails ending in '@gmail.com'.

SQL

SELECT * FROM customers WHERE email LIKE '[e-mail]';

Finds orders where the order_code has exactly three characters before '123'.

SQL

SELECT * FROM orders WHERE order_code LIKE '___123';

Sample Program

This creates a table of employees, inserts some names, and selects those whose names start with 'A'.

SQL

CREATE TABLE employees (
  id INT,
  name VARCHAR(50)
);

INSERT INTO employees (id, name) VALUES
(1, 'Alice'),
(2, 'Bob'),
(3, 'Anna'),
(4, 'Albert'),
(5, 'Brian');

SELECT * FROM employees WHERE name LIKE 'A%';

Important Notes

LIKE is case-insensitive in some databases (like MySQL) but case-sensitive in others (like PostgreSQL).

Using % at the start of the pattern can slow down the search because it cannot use indexes efficiently.

Use _ to match exactly one character when you want precise pattern matching.

Summary

WHERE with LIKE lets you search text columns using patterns.

% means any number of characters, _ means exactly one character.

It is useful for partial or flexible text matching in queries.

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'