Answers for "json decode php on place"

PHP
18

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 try json decode

/** Checks if JSON and returns decoded as an array, if not, returns false, 
but you can pass the second parameter true, if you need to return
a string in case it's not JSON */
function tryJsonDecode($string, $returnString = false) {
   $arr = json_decode($string);
  if (json_last_error() === JSON_ERROR_NONE) {
    return $arr;
  } else {
    return ($returnString) ? $string : false;
  }
}
Posted by: Guest on July-30-2021

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