Variable declaration with dollar sign in PHP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a variable named name with the value "John".

PHP

<?php
[1]name = "John";
?>

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

A$

B&

C#

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting the dollar sign before the variable name.

Using other symbols like & or # instead of $.

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

2fill in blank

medium

Complete the code to declare a variable named age with the value 25.

PHP

<?php
[1]age = 25;
?>

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

A&

B@

C#

D$

Attempts:

3 left

💡 Hint

Common Mistakes

Using symbols other than $ before the variable name.

Omitting the symbol entirely.

Variables in PHP must start with a dollar sign ($) to be valid.

3fill in blank

hard

Fix the error in the variable declaration for city with value "Paris".

PHP

<?php
[1]city = "Paris";
?>

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

Acity

B&

C$

D#

Attempts:

3 left

💡 Hint

Common Mistakes

Leaving out the dollar sign before the variable name.

Using the variable name without any symbol.

The variable must start with a dollar sign ($) to be recognized in PHP.

4fill in blank

hard

Fill both blanks to declare two variables firstName and lastName with values "Alice" and "Smith".

PHP

<?php
[1]firstName = "Alice";
[2]lastName = "Smith";
?>

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

A$

B&

C#

D@

Attempts:

3 left

💡 Hint

Common Mistakes

Using different symbols for each variable.

Omitting the dollar sign on one or both variables.

Both variables must start with a dollar sign ($) to be valid in PHP.

5fill in blank

hard

Fill all three blanks to declare variables city, country, and population with values "Rome", "Italy", and 2873000.

PHP

<?php
[1]city = "Rome";
[2]country = "Italy";
[3]population = 2873000;
?>

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

A&

B$

C#

D@

Attempts:

3 left

💡 Hint

Common Mistakes

Using different symbols for variables.

Forgetting the dollar sign before variable names.

All variables must start with a dollar sign ($) in PHP.

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