CS50 SQL: Why My `LIKE` Query Isn’t Working? (With Examples)

[Rajeev Bagra]

❓ Question

I’m working on a CS50 SQL problem using SQLite.
I have an addresses table with values like:

165 Wadsworth Place
51 Edinboro Street
251 Purchase Street

When I run this query, I either get a syntax error or no results, even though the address clearly exists:

SELECT address
FROM addresses
WHERE address LIKE '%165';

Why doesn’t this return 165 Wadsworth Place, and what is the correct way to write this query?

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✅ Answer

This is a classic SQL wildcard misunderstanding — and one of the most common mistakes beginners make in CS50.

Let’s break it down step by step.

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🧠 Understanding LIKE and % in SQL

In SQL, the % symbol is a wildcard that represents any number of characters.

But where you place % matters a lot.

Meaning of different patterns:

Pattern	What it matches
'165%'	Strings that start with 165
'%165'	Strings that end with 165
'%165%'	Strings that contain 165 anywhere

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❌ Why your query returned nothing

You wrote:

WHERE address LIKE '%165';

This means:

“Find addresses that end with 165”

But your data is:

165 Wadsworth Place

This address starts with 165 — it does not end with it.

So SQLite is correct to return no rows.

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✅ Correct query for your data

Since house numbers usually appear at the start of an address, this is the correct query:

SELECT address
FROM addresses
WHERE address LIKE '165%';

✅ Output:

165 Wadsworth Place

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❌ Another common syntax error to avoid

This query will always fail:

WHERE address LIKE %'165';

Why?

• % must be inside quotes
• There must be a space after LIKE

✅ Correct syntax:

column_name LIKE 'pattern'

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🔍 When you’re unsure of column names

Always inspect the table first:

.schema addresses

or:

PRAGMA table_info(addresses);

This prevents guessing and saves time.

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🧪 Bonus: Useful variations

Find addresses containing a street name

SELECT address
FROM addresses
WHERE address LIKE '%Wadsworth%';

Case-safe search

SELECT address
FROM addresses
WHERE UPPER(address) LIKE '165%';

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🎯 Key takeaway (CS50 exam tip)

• House numbers → start of string
• Street names → middle
• Suffixes (Street, Place, Ave) → end

So always choose your % placement based on real-world data structure, not guesswork.

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🏁 Final correct solution

SELECT address
FROM addresses
WHERE address LIKE '165%';

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