Answers for "php json_decode decode stanard class object"

PHP
20

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
0

php json_decode not working

You have to use preg_replace for avoiding the null results from json_decode

here is the example code

$json_string = stripslashes(html_entity_decode($json_string));
$bookingdata =  json_decode( preg_replace('/[x00-x1Fx80-xFF]/', '', $json_string), true );
Posted by: Guest on January-20-2022

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