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LESSON 158 OF 173 · BACKEND: BACKEND WITH AI

The AI Pasted Text into SQL in Java

Practice

An AI assistant wrote a product search for this shop: /products/search?q=te finds every product whose name contains te. It built the SQL by joining strings with +:

String sql = "SELECT name, price FROM products WHERE name LIKE '%" + q + "%'";
List<Map<String, Object>> rows = db.queryForList(sql);

The + pastes the visitor's text into the SQL itself. A quote in that text ends the SQL string early, and the rest runs as SQL: Ana's breaks the query, and x' UNION SELECT name, email FROM customers -- answers every customer's name and email, as if they were products. This is SQL injection.

The fix is a placeholder: write ? in the SQL and pass the value after it, separated by a comma. Here it is on another table:

List<Map<String, Object>> rows = db.queryForList("SELECT name FROM customers WHERE city = ?", city);

The SQL text stays the same whatever the visitor types, and the database only compares the value, quotes and all. Your code must give these answers:

GET /products/search?q=te     →  200  [{"name": "Notebook", "price": 7}, {"name": "Teapot", "price": 22}]
GET /products/search?q=MUG    →  200  [{"name": "Mug", "price": 12}]
GET /products/search?q=Ana's  →  200  []
GET /products/search?q=x' UNION SELECT name, email FROM customers --  →  200  []
GET /products/search          →  200  [{"name": "Mug", "price": 12}, {"name": "Notebook", "price": 7}, … all 8 products]

Your task: rewrite searchProducts. q arrives as its parameter ("" when it is missing). Build the pattern "%" + q + "%" as a Java string, run the query with LIKE ? and the pattern as its value, and return the rows.


Assignment
Follow the Coding Tutorial to rewrite searchProducts with a placeholder.


Hint
The % signs go into the Java string pattern, not into the SQL: the SQL only says LIKE ?. Pass pattern after the query, separated by a comma. Spring sends the list of rows as JSON.


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    Ai Sql Injection in Java

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