[Solved] Find the problem in this PHP code snippet:$items = ['x', 'y', 'z'];echo implode(' ', $items[0]); — Ans: Passing a string instead of an array to implode | PHP

PHP - String Functions

Find the problem in this PHP code snippet:

$items = ['x', 'y', 'z'];
echo implode(' ', $items[0]);

AMissing separator argument in implode

Bimplode cannot join single elements

CUsing wrong variable name

DPassing a string instead of an array to implode

Step-by-Step Solution

Solution:

Step 1: Analyze argument passed to implode
$items[0] is 'x', a string, not an array. implode expects an array.
Step 2: Confirm implode usage
implode(' ', array) is correct, but here a string is passed instead of array.
Final Answer:
Passing a string instead of an array to implode -> Option D
Quick Check:
implode requires array, not string [OK]

Quick Trick: Always pass an array to implode, not a string [OK]

Common Mistakes:

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