Variable declaration with dollar sign in PHP - Step-by-Step Execution

Concept Flow - Variable declaration with dollar sign

Start

↓

Write $variableName

↓

Assign value

↓

Use variable

↓

End

In PHP, variables always start with a dollar sign ($), then a name, then you assign a value. You can then use the variable.

Execution Sample

PHP

<?php
$age = 25;
echo $age;
?>

This code declares a variable $age with value 25 and prints it.

Execution Table

Step	Code Line	Action	Variable State	Output
1	$age = 25;	Declare $age and assign 25	$age = 25
2	echo $age;	Print value of $age	$age = 25	25
3	End	Program ends	$age = 25

💡 Program ends after printing the value of $age.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
$age	undefined	25	25	25

Key Moments - 2 Insights

Why do PHP variables always start with a dollar sign ($)?

In PHP, the dollar sign is required to tell the interpreter that this is a variable. Without it, the code will cause an error. See execution_table step 1 where $age is declared with $.

Can you use a variable without the dollar sign in PHP?

No, PHP requires the dollar sign to recognize variables. Using a name without $ is treated as a constant or string, causing errors or unexpected behavior. See execution_table step 2 where echo uses $age.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of $age after step 1?

A 25

B undefined

C 0

D null

💡 Hint

Check the 'Variable State' column in row for step 1.

At which step does the program print the value 25?

A Step 1

B Step 2

C Step 3

D No output

💡 Hint

Look at the 'Output' column in the execution table.

If you remove the $ from $age in the echo statement, what happens?

A It prints 25 anyway

B It prints the string 'age'

C It causes an error

D It prints nothing

💡 Hint

PHP treats bare words as constants; undefined ones cause an error or notice. Recall key moments.

Concept Snapshot

PHP variables always start with a $ sign.
Syntax: $variableName = value;
Use $variableName to access the value.
Without $, PHP won't recognize it as a variable.
Variables hold data you can reuse.
Example: $age = 25; echo $age;

Full Transcript

In PHP, every variable starts with a dollar sign ($). This tells PHP that what follows is a variable name. For example, $age = 25; declares a variable named age with the value 25. When you want to use the variable, you must include the $ sign, like echo $age; which prints 25. If you forget the $, PHP will not understand it as a variable and will cause an error. This simple rule helps PHP know what is a variable and what is not.

Practice

1. Which of the following is the correct way to declare a variable named age in PHP?

easy

A. $age = 25;

B. age = 25;

C. var age = 25;

D. int $age = 25;

Solution

1. Step 1: Recognize PHP variable syntax

   In PHP, all variables must start with a dollar sign ($) followed by the variable name.

2. Step 2: Check each option for correct syntax

   $age = 25; uses $age = 25; which is correct. Options B and C miss the dollar sign, and A uses a type declaration which is not standard in PHP variable declaration.

3. Final Answer:

   $age = 25; -> Option A

4. Quick Check:

   PHP variables start with $ = $age = 25; [OK]

Hint: Always start PHP variables with $ sign [OK]

Common Mistakes:

• Forgetting the $ sign before variable name
• Using other language syntax like var or int
• Trying to declare type explicitly in PHP

2. Which of the following is a syntax error in PHP variable declaration?

easy

A. $1number = 10;

B. $name = "John";

C. $total = 100;

D. $_score = 50;

Solution

1. Step 1: Understand valid variable names in PHP

   Variable names must start with a letter or underscore, not a number.

2. Step 2: Identify invalid variable name

   $1number = 10; uses $1number which starts with a number, causing a syntax error. Other options start with $ followed by letter or underscore, which is valid.

3. Final Answer:

   $1number = 10; -> Option A

4. Quick Check:

   Variable names cannot start with digits = $1number = 10; [OK]

Hint: Variable names cannot start with numbers [OK]

Common Mistakes:

• Starting variable names with digits
• Using spaces in variable names
• Using special characters other than underscore

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

$x = 5;
$y = $x;
$x = 10;
echo $y;

medium

A. Error

B. 10

C. 0

D. 5

Solution

1. Step 1: Understand variable assignment in PHP

   When $y = $x; runs, $y gets the current value of $x, which is 5.

2. Step 2: Check effect of changing $x after assignment

   Changing $x to 10 later does not affect $y because $y holds a copy of the value, not a reference.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Variable copy keeps old value = 5 [OK]

Hint: Variable assignment copies value, not reference [OK]

Common Mistakes:

• Assuming $y changes when $x changes
• Confusing references with copies
• Expecting runtime error for reassignment

4. Identify the error in this PHP code snippet:

$var1 = 10;
var2 = 20;
echo $var2;

medium

A. echo statement syntax error

B. Missing $ sign before var2

C. Missing semicolon after var1

D. Variable names cannot have numbers

Solution

1. Step 1: Check variable declarations

   $var1 is declared correctly with a $ sign. var2 is missing the $ sign, which is required in PHP.

2. Step 2: Analyze the echo statement

   echo uses $var2 which is undefined because var2 was not declared properly.

3. Final Answer:

   Missing $ sign before var2 -> Option B

4. Quick Check:

   All PHP variables need $ sign = Missing $ sign before var2 [OK]

Hint: Every PHP variable must start with $ [OK]

Common Mistakes:

• Forgetting $ sign before variable
• Assuming semicolon missing causes error here
• Thinking numbers are not allowed in variable names

5. You want to store a user's first name and age in PHP variables and then display them. Which code correctly declares and uses these variables?

1. $firstName = "Alice";
2. $age = 30;
3. echo "Name: $firstName, Age: $age";

hard

A. Declare variables without $ sign and use $ in echo

B. Use variables without $ sign in echo: echo "Name: firstName, Age: age";

C. Declare variables with $ and use them with $ in echo as shown

D. Use $ only in declaration, not in echo

Solution

1. Step 1: Declare variables with $ sign

   Lines 1 and 2 correctly declare variables $firstName and $age with $ sign.

2. Step 2: Use variables with $ sign inside double quotes

   In PHP, variables inside double quotes are parsed and replaced with their values, so echo "Name: $firstName, Age: $age"; works correctly.

3. Final Answer:

   Declare variables with $ and use them with $ in echo as shown -> Option C

4. Quick Check:

   Variables need $ in declaration and usage [OK]

Hint: Use $ sign in both declaration and usage [OK]

Common Mistakes:

• Omitting $ sign in echo
• Using single quotes which don't parse variables
• Declaring variables without $ sign