Global keyword behavior in PHP - Mini Project: Build & Apply

Global keyword behavior

📖 Scenario: Imagine you have a small PHP script that keeps track of a counter. You want to increase this counter inside a function, but the counter is defined outside the function.

🎯 Goal: You will learn how to use the global keyword in PHP to access and modify a variable defined outside a function from inside that function.

📋 What You'll Learn

Create a variable called $counter with the value 0 outside any function.

Create a variable called $increment with the value 5.

Create a function called increaseCounter that uses the global keyword to access $counter and adds $increment to it.

Call the function increaseCounter once.

Print the value of $counter.

💡 Why This Matters

🌍 Real World

Using the <code>global</code> keyword helps when you want to share and update data across different parts of a PHP script, like counters, settings, or shared resources.

💼 Career

Understanding variable scope and the <code>global</code> keyword is important for PHP developers to manage data correctly and avoid bugs in web applications.

1

Create the initial counter variable

Create a variable called $counter and set it to 0.

PHP

$counter = 0;
# Your code here

Hint

Use $counter = 0; to create the variable.

2

Add the increment variable

Create a variable called $increment and set it to 5.

PHP

<?php
$counter = 0;
$increment = 5;
# Your code here

Hint

Use $increment = 5; to create the variable.

3

Create the function to increase the counter

Create a function called increaseCounter that uses the global keyword to access $counter and adds $increment to it.

PHP

<?php
$counter = 0;
$increment = 5;

function increaseCounter() {
    // Your code here
}

Hint

Inside the function, write global $counter, $increment; to access the variables, then add $increment to $counter.

4

Call the function and print the counter

Call the function increaseCounter() once, then print the value of $counter using echo.

PHP

<?php
$counter = 0;
$increment = 5;

function increaseCounter() {
    global $counter, $increment;
    $counter += $increment;
}

// Your code here

Hint

Call increaseCounter(); then print $counter with echo $counter;.

Practice

1. What does the global keyword do in PHP?

easy

A. It deletes a variable from the global scope.

B. It creates a new local variable inside the function.

C. It makes a variable constant and unchangeable.

D. It allows a function to access variables defined outside its scope.

Solution

1. Step 1: Understand variable scopes in PHP

   Variables defined outside functions are in the global scope and not accessible inside functions by default.

2. Step 2: Role of the global keyword

   The global keyword inside a function tells PHP to use the variable from the global scope instead of creating a new local one.

3. Final Answer:

   It allows a function to access variables defined outside its scope. -> Option D

4. Quick Check:

   global keyword = access outside variables [OK]

Hint: Global lets functions use outside variables directly [OK]

Common Mistakes:

• Thinking global creates a new variable inside the function
• Assuming global deletes variables
• Confusing global with constants

2. Which of the following is the correct way to use the global keyword inside a function to access a variable $count defined outside?

easy

A. global $count;

B. use $count;

C. global($count);

D. extern $count;

Solution

1. Step 1: Recall PHP syntax for global variables

   In PHP, to access a global variable inside a function, you write global $variableName; without parentheses.

2. Step 2: Check each option

   global $count; uses correct syntax. use $count; is invalid in PHP. global($count); uses parentheses which is incorrect. extern $count; uses 'extern' which is not a PHP keyword.

3. Final Answer:

   global $count; -> Option A

4. Quick Check:

   Correct syntax = global $var; [OK]

Hint: Use 'global $var;' without parentheses inside functions [OK]

Common Mistakes:

• Adding parentheses after global
• Using 'use' or 'extern' keywords incorrectly
• Forgetting the $ sign before variable name

3. What will be the output of the following PHP code?

$x = 5;
function test() {
  global $x;
  $x = $x + 10;
}
test();
echo $x;

medium

A. 15

B. 5

C. 10

D. Error

Solution

1. Step 1: Understand the initial value of $x

   $x is set to 5 in the global scope before the function call.

2. Step 2: Effect of the function test()

   Inside test(), global $x; makes $x inside the function refer to the global $x. Then $x is increased by 10, so $x becomes 15 globally.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   global $x modified to 15 [OK]

Hint: global lets function change outside variable value [OK]

Common Mistakes:

• Thinking $x inside function is local and unchanged
• Expecting output 5 or 10 instead of 15
• Assuming error due to global keyword

4. Identify the error in this PHP code snippet:

$count = 10;
function increment() {
  $count = $count + 1;
}
increment();
echo $count;

medium

A. Variable $count is undefined.

B. Syntax error in function declaration.

C. Missing global keyword inside function to access $count.

D. Cannot echo variable outside function.

Solution

1. Step 1: Check variable scope inside function

   Inside increment(), $count is treated as a new local variable, not the global one.

2. Step 2: Missing global keyword

   To modify the global $count, the function must declare global $count; before using it.

3. Final Answer:

   Missing global keyword inside function to access $count. -> Option C

4. Quick Check:

   global needed to access outside variable [OK]

Hint: Declare global inside function to modify outside variable [OK]

Common Mistakes:

• Assuming $count inside function is global by default
• Thinking echo outside function causes error
• Confusing syntax errors with scope errors

5. Given the code below, what will be the output?

$a = 3;
$b = 4;
function sum() {
  global $a, $b;
  $b = $a + $b;
}
sum();
echo $b;

What if global $a, $b; is replaced with global $a; only?

hard

A. Output is 7 in both cases.

B. Output is 7 with global $a, $b; and 4 with global $a; only.

C. Output is 4 with global $a, $b; and 7 with global $a; only.

D. Code causes error without global $b;.

Solution

1. Step 1: When using global $a, $b;

   Both $a and $b inside the function refer to global variables. $b is updated to 3 + 4 = 7 globally.

2. Step 2: When using global $a; only

   $a inside function is global, but $b is local and uninitialized. So $b = $a + $b uses local $b (null or 0), $b becomes 3 + 0 = 3 locally, global $b remains 4. echo $b outputs global $b which is 4.

3. Final Answer:

   Output is 7 with global $a, $b; and 4 with global $a; only. -> Option B

4. Quick Check:

   global vars must be declared to modify them [OK]

Hint: Declare all globals you want to use inside function [OK]

Common Mistakes:

• Assuming global $a; also makes $b global
• Expecting output 7 without declaring global $b
• Thinking missing global causes syntax error