How SQL injection exploits unsafe queries in PHP

Introduction

SQL injection happens when bad users put harmful commands into places where your program expects normal data. This can let them see or change your database without permission.

When you accept user input to search or filter data in a database.

When you build SQL queries by adding user input directly into the query string.

When you want to understand why your website or app might be unsafe.

When you want to learn how to protect your database from attackers.

Syntax

PHP

<?php
// Unsafe SQL query example
$user_input = $_GET['id'];
$query = "SELECT * FROM users WHERE id = '$user_input'";
$result = mysqli_query($conn, $query);
?>

This example shows how user input is directly added to the SQL query.

This is unsafe because a user can add extra commands to the input.

Examples

User input is added directly to the query string, which is unsafe.

PHP

<?php
// Unsafe example
$id = $_GET['id'];
$query = "SELECT * FROM users WHERE id = '$id'";
mysqli_query($conn, $query);

This uses a prepared statement to safely insert user input.

PHP

<?php
// Safe example using prepared statements
$id = $_GET['id'];
$stmt = $conn->prepare("SELECT * FROM users WHERE id = ?");
$stmt->bind_param("i", $id);
$stmt->execute();

Sample Program

This program shows how a user input like "1 OR 1=1" can make the query return all rows, not just one.

PHP

<?php
// Example showing SQL injection vulnerability
$conn = new mysqli('localhost', 'user', 'pass', 'testdb');

// Simulate user input that tries SQL injection
$user_input = "1 OR 1=1";

// Unsafe query
$query = "SELECT * FROM users WHERE id = '$user_input'";

// Run query
$result = $conn->query($query);

// Print number of rows returned
if ($result) {
    echo "Number of rows: " . $result->num_rows . "\n";
} else {
    echo "Query error";
}
?>

Important Notes

Never put user input directly into SQL queries.

Use prepared statements or parameterized queries to keep your database safe.

SQL injection can let attackers see or change data they should not access.

Summary

SQL injection happens when unsafe queries let users add harmful commands.

Always use safe methods like prepared statements to handle user input.

Protecting your database keeps your app and users safe.

Practice

1. What is the main risk of using unsafe SQL queries in PHP without proper input handling?

easy

A. The website will load faster

B. Users will see more colorful pages

C. Attackers can insert harmful SQL commands to access or damage the database

D. The server will automatically block all users

Solution

1. Step 1: Understand unsafe SQL queries

   Unsafe queries allow user input directly in SQL commands without checks.

2. Step 2: Identify the risk of SQL injection

   This lets attackers add harmful commands to steal or change data.

3. Final Answer:

   Attackers can insert harmful SQL commands to access or damage the database -> Option C

4. Quick Check:

   Unsafe queries = SQL injection risk [OK]

Hint: Unsafe queries let attackers add harmful commands [OK]

Common Mistakes:

• Thinking unsafe queries improve speed
• Confusing SQL injection with visual changes
• Assuming server blocks all users automatically

2. Which PHP code snippet is unsafe and vulnerable to SQL injection?

easy

A. $query = "SELECT * FROM users WHERE name = '" . $_GET['name'] . "'";

B. $stmt = $pdo->prepare('SELECT * FROM users WHERE name = ?'); $stmt->execute([$_GET['name']]);

C. $stmt = $pdo->prepare('SELECT * FROM users WHERE name = :name'); $stmt->execute(['name' => $_GET['name']]);

D. $query = 'SELECT * FROM users WHERE name = :name'; $stmt = $pdo->prepare($query); $stmt->execute(['name' => 'Alice']);

Solution

1. Step 1: Identify direct user input in query string

   $query = "SELECT * FROM users WHERE name = '" . $_GET['name'] . "'"; directly inserts $_GET['name'] into the SQL string without protection.

2. Step 2: Compare with prepared statements

   Options A, B, and D use prepared statements which safely handle input.

3. Final Answer:

   $query = "SELECT * FROM users WHERE name = '" . $_GET['name'] . "'"; -> Option A

4. Quick Check:

   Direct input in query = unsafe [OK]

Hint: Directly adding user input to query is unsafe [OK]

Common Mistakes:

• Confusing prepared statements with unsafe queries
• Thinking all string concatenations are safe
• Ignoring the use of placeholders in queries

3. What will be the output of this PHP code if the URL parameter is ?id=1 OR 1=1?

$id = $_GET['id'];
$query = "SELECT * FROM users WHERE id = $id";
echo $query;

medium

A. SELECT * FROM users WHERE id = 1

B. SELECT * FROM users WHERE id = 1 OR 1=1

C. Syntax error in SQL query

D. SELECT * FROM users WHERE id = '1 OR 1=1'

Solution

1. Step 1: Substitute the URL parameter into the query

   The variable $id becomes the string "1 OR 1=1" directly inserted into the query.

2. Step 2: Form the final SQL query string

   The query becomes: SELECT * FROM users WHERE id = 1 OR 1=1, which is a valid SQL statement.

3. Final Answer:

   SELECT * FROM users WHERE id = 1 OR 1=1 -> Option B

4. Quick Check:

   Unsafe input inserted directly = query with injected condition [OK]

Hint: Direct input inserts raw text into query [OK]

Common Mistakes:

• Assuming input is automatically quoted
• Expecting syntax error without quotes
• Ignoring how SQL interprets OR conditions

4. Identify the error in this PHP code that makes it vulnerable to SQL injection:

$username = $_POST['username'];
$query = "SELECT * FROM users WHERE username = '$username'";
$result = mysqli_query($conn, $query);

medium

A. The query string is missing a semicolon

B. The variable $username is not defined

C. mysqli_query cannot run SELECT queries

D. Using prepared statements would prevent SQL injection

Solution

1. Step 1: Check how user input is used in the query

   The variable $username is inserted directly into the SQL string without sanitization.

2. Step 2: Suggest the fix to prevent injection

   Using prepared statements safely handles input and prevents injection attacks.

3. Final Answer:

   Using prepared statements would prevent SQL injection -> Option D

4. Quick Check:

   Direct input in query = vulnerability [OK]

Hint: Use prepared statements to fix injection bugs [OK]

Common Mistakes:

• Thinking variable is undefined
• Believing mysqli_query can't run SELECT
• Focusing on missing semicolon instead of security

5. You want to fix this vulnerable PHP code to prevent SQL injection:

$email = $_POST['email'];
$query = "SELECT * FROM users WHERE email = '$email'";
$result = mysqli_query($conn, $query);

Which is the best way to fix it?

hard

A. $stmt = $conn->prepare('SELECT * FROM users WHERE email = ?'); $stmt->bind_param('s', $email); $stmt->execute();

B. $email = addslashes($_POST['email']); $query = "SELECT * FROM users WHERE email = '$email'"; mysqli_query($conn, $query);

C. $query = "SELECT * FROM users WHERE email = '" . $_POST['email'] . "'"; mysqli_query($conn, $query);

D. $email = htmlspecialchars($_POST['email']); $query = "SELECT * FROM users WHERE email = '$email'"; mysqli_query($conn, $query);

Solution

1. Step 1: Review each option's security

   $stmt = $conn->prepare('SELECT * FROM users WHERE email = ?'); $stmt->bind_param('s', $email); $stmt->execute(); uses prepared statements with parameter binding, which safely handles input.

2. Step 2: Compare other options

   $email = addslashes($_POST['email']); $query = "SELECT * FROM users WHERE email = '$email'"; mysqli_query($conn, $query); uses addslashes which is not fully safe; C is unsafe direct insertion; D uses htmlspecialchars which is for HTML, not SQL.

3. Final Answer:

   $stmt = $conn->prepare('SELECT * FROM users WHERE email = ?'); $stmt->bind_param('s', $email); $stmt->execute(); -> Option A

4. Quick Check:

   Prepared statements = best SQL injection protection [OK]

Hint: Use prepared statements with bind_param for safety [OK]

Common Mistakes:

• Relying on addslashes for security
• Using htmlspecialchars for SQL input
• Continuing direct string concatenation