[Solved] Identify the error in this PHP code snippet:try { throw new Exception('Error!'); } catch (Error $e) { echo 'Caught Error'; } | PHP

PHP - Error and Exception Handling

Identify the error in this PHP code snippet:

try {
  throw new Exception('Error!');
} catch (Error $e) {
  echo 'Caught Error';
}

AThe catch block must come before try block.

BThe throw statement is missing a semicolon.

CThe try block cannot throw exceptions.

DThe catch block should catch Exception, not Error.

Step-by-Step Solution

Solution:

Step 1: Check thrown object type
The code throws an Exception object.
Step 2: Check catch block type
The catch block is for Error, which does not catch Exception.
Final Answer:
The catch block should catch Exception, not Error. -> Option D
Quick Check:
Catch type must match thrown exception [OK]

Quick Trick: Catch block type must match thrown exception type [OK]

Common Mistakes:

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