[Solved] What will be the output of this PHP code?$json = '{"name":"John","age":30}';$obj = json_decode($json);echo $obj->name; | PHP

PHP - File Handling

What will be the output of this PHP code?

$json = '{"name":"John","age":30}';
$obj = json_decode($json);
echo $obj->name;

Aname

BError: Cannot access property

CArray

DJohn

Step-by-Step Solution

Solution:

Step 1: Decode JSON string to object
json_decode without second argument returns an object with properties matching JSON keys.
Step 2: Access object property
Accessing $obj->name returns the string "John".
Final Answer:
John -> Option D
Quick Check:
Access object property after json_decode = value [OK]

Quick Trick: Access object properties with -> after json_decode [OK]

Common Mistakes:

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