Variable declaration with dollar sign in PHP - Mini Project: Build & Apply

Variable declaration with dollar sign

📖 Scenario: Imagine you are creating a simple PHP script to store and display information about a fruit.

🎯 Goal: You will declare variables using the dollar sign, assign values, and then display the values.

📋 What You'll Learn

Declare variables with the dollar sign ($)

Assign string and integer values to variables

Use echo to display variable values

💡 Why This Matters

🌍 Real World

Variables with dollar signs are the basic way to store and use data in PHP scripts, such as in websites or apps.

💼 Career

Understanding variable declaration is essential for any PHP developer to write dynamic and interactive web pages.

1

Declare variables with dollar sign

Create a variable called $fruit and assign it the value "Apple". Also create a variable called $quantity and assign it the value 5.

PHP

<?php

# Declare the variables $fruit and $quantity with the given values

Hint

Remember, in PHP, all variables start with a dollar sign ($).

2

Add a variable for price

Create a variable called $price and assign it the value 2.5.

PHP

<?php
$fruit = "Apple";
$quantity = 5;

# Declare the variable $price with the value 2.5

Hint

Use the dollar sign ($) before the variable name and assign the value with =.

3

Calculate total cost

Create a variable called $total and assign it the value of $quantity multiplied by $price.

PHP

<?php
$fruit = "Apple";
$quantity = 5;
$price = 2.5;

# Calculate $total as $quantity times $price

Hint

Use the * operator to multiply variables.

4

Display the total cost

Use echo to display the text: Total cost for 5 Apples is 12.5 using the variables $quantity, $fruit, and $total.

PHP

<?php
$fruit = "Apple";
$quantity = 5;
$price = 2.5;
$total = $quantity * $price;

# Display the total cost using echo and variables

Hint

Use double quotes "" to include variables inside the string with echo.

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