Answers for "php try to decode json"

PHP
19

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 and check

// Checks if json
function isJson($string) {
   json_decode($string);
   return json_last_error() === JSON_ERROR_NONE;
}

// example
if (isJson($string) {
  // Do your stuff here
}
Posted by: Guest on July-30-2021
0

php json decode

$obj = json_decode("{string:'string'}");
Posted by: Guest on March-08-2021
0

php try to decode json

/** 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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