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PHPprogramming~3 mins

How SQL injection exploits unsafe queries in PHP - Why You Should Know This

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The Big Idea

What if a simple login box could open the door to your entire database?

The Scenario

Imagine you have a website where users type their username and password to log in. You write code that directly adds what users type into your database query without checking it first.

At first, it seems simple and works fine for normal users.

The Problem

But what if someone types strange code instead of a username? Because your code just adds their input directly, this can trick your database into running commands you never wanted.

This can let bad people see private data or even change your database without permission.

The Solution

By understanding how SQL injection works, you learn to write safer code that keeps user input separate from commands.

This stops attackers from tricking your database and keeps your data safe.

Before vs After
✗ Before
$query = "SELECT * FROM users WHERE username = '" . $_POST['username'] . "' AND password = '" . $_POST['password'] . "'";
✓ After
$stmt = $pdo->prepare('SELECT * FROM users WHERE username = ? AND password = ?');
$stmt->execute([$_POST['username'], $_POST['password']]);
What It Enables

It enables you to build secure applications that protect user data and prevent attackers from breaking in.

Real Life Example

Many websites have been hacked because they used unsafe queries. Learning this helps you avoid those mistakes and keep your site safe.

Key Takeaways

Unsafe queries let attackers run harmful commands.

SQL injection happens when user input is not handled safely.

Using safe query methods protects your database and users.

Practice

(1/5)
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
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