String interpolation in double quotes in PHP - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print the variable inside double quotes using interpolation.

PHP

<?php
$name = "Alice";
echo "Hello, [1]!";
?>

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

Aname

B$name

C'$name'

D{name}

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting the $ sign before the variable name.

Using single quotes which do not interpolate variables.

In PHP, variables inside double quotes are interpolated when prefixed with $. So "$name" prints the variable's value.

2fill in blank

medium

Complete the code to include the variable inside a sentence using double quotes and interpolation.

PHP

<?php
$age = 25;
echo "I am [1] years old.";
?>

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

A$age

B'$age'

Cage

D{age}

Attempts:

3 left

💡 Hint

Common Mistakes

Using single quotes which do not interpolate variables.

Omitting the $ sign before the variable.

Using $age inside double quotes prints the value of the variable age.

3fill in blank

hard

Fix the error in the code to correctly interpolate the variable inside double quotes.

PHP

<?php
$city = "Paris";
echo "Welcome to [1]!";
?>

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

A{city}

Bcity

C'$city'

D$city

Attempts:

3 left

💡 Hint

Common Mistakes

Using the variable name without $ inside double quotes.

Using single quotes which do not interpolate variables.

Variables inside double quotes must have the $ sign to be interpolated correctly.

4fill in blank

hard

Fill both blanks to correctly interpolate the variable and add a suffix inside double quotes.

PHP

<?php
$score = 90;
echo "Your score is [1][2].";
?>

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

A$score

B%

C points

Dscore

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting the $ sign before the variable.

Using % instead of ' points' as suffix.

Use $score to interpolate the variable and add ' points' as a suffix inside the string.

5fill in blank

hard

Fill all three blanks to create a sentence with two interpolated variables and a connector inside double quotes.

PHP

<?php
$firstName = "John";
$lastName = "Doe";
echo "My name is [1] [2] [3].";
?>

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

A$firstName

B$lastName

Cand

Dor

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names without $ sign.

Using 'or' instead of 'and' as connector.

Use $firstName and $lastName to interpolate the variables and 'and' as the connector word.

Practice

1. What does string interpolation in PHP allow you to do inside double quotes?

easy

A. Execute functions inside the string

B. Insert variable values directly inside the string

C. Create arrays inside the string

D. Add comments inside the string

Solution

1. Step 1: Understand string interpolation

   String interpolation means putting variables inside double-quoted strings so their values appear directly.

2. Step 2: Identify what interpolation does

   It replaces variable names with their values inside the string automatically.

3. Final Answer:

   Insert variable values directly inside the string -> Option B

4. Quick Check:

   Interpolation = Insert variable values [OK]

Hint: Double quotes allow variables inside strings [OK]

Common Mistakes:

• Thinking interpolation runs functions
• Believing arrays can be created inside strings
• Confusing comments with interpolation

2. Which of the following is the correct way to interpolate a variable $name inside a string in PHP?

easy

A. echo 'Hello, $name!';

B. echo Hello, $name!;

C. echo "Hello, '$name'!";

D. echo "Hello, $name!";

Solution

1. Step 1: Check string delimiters for interpolation

   PHP interpolates variables only inside double quotes, not single quotes.

2. Step 2: Analyze each option

   echo "Hello, $name!"; uses double quotes and variable $name correctly inside the string.

3. Final Answer:

   echo "Hello, $name!"; -> Option D

4. Quick Check:

   Double quotes + $variable = interpolation [OK]

Hint: Use double quotes for variables inside strings [OK]

Common Mistakes:

• Using single quotes which do not interpolate
• Omitting quotes around the string
• Putting variable inside single quotes inside double quotes

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

$fruit = "apple";
echo "I like $fruit pie.";

medium

A. I like apple pie.

B. I like $fruit pie.

C. I like $fruitpie.

D. Syntax error

Solution

1. Step 1: Understand variable interpolation in double quotes

   The variable $fruit inside double quotes will be replaced by its value "apple".

2. Step 2: Check the output string

   The string becomes "I like apple pie." and is printed as is.

3. Final Answer:

   I like apple pie. -> Option A

4. Quick Check:

   Variable replaced by value = I like apple pie. [OK]

Hint: Variables inside " " show their values [OK]

Common Mistakes:

• Expecting variable name printed literally
• Confusing variable concatenation without braces
• Thinking it causes syntax error

4. Identify the error in this PHP code snippet:

$color = "red";
echo "The color is $colorish.";

medium

A. Outputs 'The color is ' with no variable value

B. Syntax error due to missing semicolon

C. Outputs 'The color is ' followed by undefined variable error

D. No error, outputs 'The color is redish.'

Solution

1. Step 1: Analyze variable name in string

   PHP tries to find variable named $colorish, but only $color exists.

2. Step 2: Understand variable interpolation rules

   Without braces, PHP treats $colorish as one variable, causing undefined variable notice.

3. Step 3: Fix by using braces

   Using "{$color}ish" would correctly output "redish".

4. Final Answer:

   Outputs 'The color is ' followed by undefined variable error -> Option C

5. Quick Check:

   Variable name confusion = undefined variable error [OK]

Hint: Use braces {} to separate variable names from text [OK]

Common Mistakes:

• Assuming $colorish is $color + 'ish'
• Ignoring undefined variable warnings
• Not using braces for complex interpolation

5. Given the variables $user = 'Alice' and $count = 5, which code correctly outputs:
Hello Alice, you have 5 new messages. using string interpolation?

hard

A. echo "Hello {$user}, you have {$count} new messages.";

B. echo 'Hello $user, you have $count new messages.';

C. echo "Hello $user, you have {$count new messages}.";

D. echo "Hello $user, you have $countnew messages.";

Solution

1. Step 1: Check variable interpolation with and without braces

   Braces are needed when a variable is followed by alphanumeric characters that could extend the variable name (e.g., $countnew).

2. Step 2: Analyze each option

   A: echo "Hello $user, you have $countnew messages."; treats $countnew as undefined variable (wrong output). B: Single quotes (no interpolation). C: Syntax error inside braces. D: Braces used correctly for safe interpolation.

3. Final Answer:

   echo "Hello {$user}, you have {$count} new messages."; -> Option A

4. Quick Check:

   Braces clarify variables inside strings [OK]

Hint: Use braces {} around variables for safe interpolation [OK]

Common Mistakes:

• Using single quotes which prevent interpolation
• Incorrect brace placement causing syntax errors
• Omitting braces when needed for clarity