[Solved] Identify the error in this PHP code snippet:$json = '{"fruit":"apple","count":3}'; $data = json_decode($json, true); echo $data->fruit; | PHP

PHP - File Handling

Identify the error in this PHP code snippet:

$json = '{"fruit":"apple","count":3}';
$data = json_decode($json, true);
echo $data->fruit;

Ajson_decode cannot parse this JSON

BIncorrect JSON string format

CMissing second parameter in json_decode

DUsing object syntax on an array after json_decode

Step-by-Step Solution

Solution:

Step 1: Check json_decode parameters
The second parameter true makes json_decode return an associative array.
Step 2: Accessing array with object syntax
Using $data->fruit tries to access an object property, but $data is an array, so this causes an error.
Final Answer:
Using object syntax on an array after json_decode -> Option D
Quick Check:
json_decode(true) returns array, use $data['fruit'] [OK]

Quick Trick: Use $data['key'] if json_decode(true) used [OK]

Common Mistakes:

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