[Solved] Examine the following PHP code snippet:$var = nullif ($var === null) { echo 'Null value';}What is the issue with this code? | PHP

PHP - Variables and Data Types

Examine the following PHP code snippet:

$var = null
if ($var === null) {
  echo 'Null value';
}

What is the issue with this code?

AMissing semicolon after <code>$var = null</code>

BUsing strict comparison operator <code>===</code> incorrectly

CEcho statement syntax is invalid

DVariable <code>$var</code> is not declared

Step-by-Step Solution

Solution:

Step 1: Check PHP statement syntax
Every PHP statement must end with a semicolon (;).
Step 2: Identify missing semicolon
The line $var = null lacks a semicolon at the end, causing a syntax error.
Step 3: Verify other parts
The strict comparison === is valid, echo syntax is correct, and $var is declared.
Final Answer:
Missing semicolon after $var = null -> Option A
Quick Check:
Statements must end with ';' [OK]

Quick Trick: Always end PHP statements with a semicolon [OK]

Common Mistakes:

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