Answers for "json decode php new var"

php json_decode

$personJSON = '{"name":"Johny Carson","title":"CTO"}';

$person = json_decode($personJSON);

echo $person->name; // Johny Carson

Posted by: Guest on July-01-2019

php json decode

$obj = json_decode("{string:'string'}");

Posted by: Guest on March-08-2021

Code answers related to "json decode php new var"