How SQL injection exploits unsafe queries in PHP - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the user input safely.

PHP

<?php
$username = $_GET[[1]];
?>

Drag options to blanks, or click blank then click option'

Ausername

B"user"

C'user'

Duser

Attempts:

3 left

💡 Hint

Common Mistakes

Not using quotes around the key name causes a syntax error.

Using single quotes inside single quotes without escaping.

The user input is accessed via the $_GET superglobal with the key as a string, so it must be quoted.

2fill in blank

medium

Complete the code to build an unsafe SQL query using the user input.

PHP

<?php
$query = "SELECT * FROM users WHERE username = '[1]'";
?>

Drag options to blanks, or click blank then click option'

A$username

Busername

C'$username'

D"$username"

Attempts:

3 left

💡 Hint

Common Mistakes

Using the variable name without $ sign.

Adding extra quotes around the variable causing syntax errors.

The variable $username should be inserted directly inside the string with single quotes around it in the SQL query.

3fill in blank

hard

Fix the error in the unsafe query that allows SQL injection.

PHP

<?php
$query = "SELECT * FROM users WHERE username = '" . [1] . "'";
?>

Drag options to blanks, or click blank then click option'

Amysqli_real_escape_string($conn, $username)

B$username

Caddslashes($username)

Dhtmlspecialchars($username)

Attempts:

3 left

💡 Hint

Common Mistakes

Using htmlspecialchars which is for HTML, not SQL.

Not escaping the input at all.

Using mysqli_real_escape_string escapes special characters to prevent SQL injection.

4fill in blank

hard

Fill both blanks to prepare and execute a safe SQL statement.

PHP

<?php
$stmt = $conn->prepare("SELECT * FROM users WHERE username = [1]");
$stmt->bind_param([2], $username);
$stmt->execute();
?>

Drag options to blanks, or click blank then click option'

A?

B"s"

Cs

D'?'

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around the placeholder or type causes errors.

Using the wrong placeholder symbol.

The placeholder in the SQL statement is a question mark ?. The bind_param type for a string is s without quotes.

5fill in blank

hard

Fill all three blanks to fetch and display safe query results.

PHP

<?php
$stmt = $conn->prepare("SELECT email FROM users WHERE username = [1]");
$stmt->bind_param([2], $username);
$stmt->bind_result([3]);
$stmt->execute();
$stmt->fetch();
echo $email;
?>

Drag options to blanks, or click blank then click option'

A?

Bs

C$email

D"s"

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around the bind_param type.

Not using a variable with $ in bind_result.

The placeholder is ?, the bind_param type is s, and the variable to bind the result is $email.

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