WHERE with LIKE pattern matching in SQL - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ query_result

intermediate

Find names starting with 'Jo' using LIKE

Given a table Employees with a column name, which query returns all employees whose names start with 'Jo'?

SQL

SELECT name FROM Employees WHERE name LIKE 'Jo%';

A SELECT name FROM Employees WHERE name LIKE '%Jo%';

B SELECT name FROM Employees WHERE name LIKE '%Jo';

C SELECT name FROM Employees WHERE name LIKE 'J_o%';

D SELECT name FROM Employees WHERE name LIKE 'Jo%';

💡 Hint

The percent sign (%) matches any sequence of characters, and placing it after 'Jo' means names starting with 'Jo'.

Option D correctly uses 'Jo%' to match any name starting with 'Jo'. Option D matches names ending with 'Jo'. Option D matches names containing 'Jo' anywhere. Option D matches names starting with 'J' followed by any single character, then 'o'.

❓ query_result

intermediate

Select emails containing 'admin' anywhere

Which SQL query returns all rows from Users where the email contains the word 'admin' anywhere in the string?

SQL

SELECT email FROM Users WHERE email LIKE '%admin%';

A SELECT email FROM Users WHERE email LIKE '%admin';

B SELECT email FROM Users WHERE email LIKE '%admin%';

C SELECT email FROM Users WHERE email LIKE 'admin%';

D SELECT email FROM Users WHERE email LIKE '_admin%';

💡 Hint

Use % before and after 'admin' to find it anywhere in the string.

Option B uses '%admin%' which matches any string containing 'admin' anywhere. Option B matches strings starting with 'admin'. Option B matches strings ending with 'admin'. Option B matches strings with any single character before 'admin' at the start.

📝 Syntax

advanced

Identify the query with syntax error in LIKE pattern

Which SQL query will cause a syntax error due to incorrect LIKE pattern?

A SELECT * FROM Orders WHERE order_id LIKE '12%3_'

B SELECT * FROM Orders WHERE order_id LIKE '123%';

C SELECT * FROM Orders WHERE order_id LIKE '12_3%';

D SELECT * FROM Orders WHERE order_id LIKE '%123';

💡 Hint

Check for missing quotes or incomplete strings in the LIKE pattern.

Option A is missing the closing quote for the LIKE pattern, causing a syntax error. All other options have correct syntax.

❓ optimization

advanced

Optimize query using LIKE for prefix search

You want to find all customers whose city starts with 'San'. Which query is most efficient for this search?

A SELECT * FROM Customers WHERE city LIKE '%San';

B SELECT * FROM Customers WHERE city LIKE '%San%';

C SELECT * FROM Customers WHERE city LIKE 'San%';

D SELECT * FROM Customers WHERE city LIKE '_San%';

💡 Hint

Using a pattern that starts with the search string allows index use.

Option C uses 'San%' which matches cities starting with 'San' and can use indexes efficiently. Option C and C use leading %, which prevents index use. Option C matches cities with any single character before 'San', which is not the intended search.

🧠 Conceptual

expert

Understanding LIKE pattern matching with escape characters

Given a table Files with a column filename, which query correctly finds all files with names ending with '_backup' (underscore is a literal character, not a wildcard)?

A SELECT filename FROM Files WHERE filename LIKE '%\_backup' ESCAPE '\';

B SELECT filename FROM Files WHERE filename LIKE '%[_]backup';

C SELECT filename FROM Files WHERE filename LIKE '%_backup';

D SELECT filename FROM Files WHERE filename LIKE '%\_backup';

💡 Hint

Use ESCAPE clause to treat underscore as a normal character.

Option A correctly escapes the underscore using '\' and specifies ESCAPE '\' so underscore is treated literally. Option A escapes underscore but lacks ESCAPE clause, so backslash is treated literally. Option A treats underscore as wildcard matching any single character. Option A uses character class which may work in some SQL dialects but is not standard.

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'