Answers for "php json_decode error check"

php json_decode

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

$person = json_decode($personJSON);

echo $person->name; // Johny Carson

php decode json object

<?php

$json = '{"firstName":"Peter","lastName:":"Silva","age":23}';

$personInfo = json_decode(json);

echo $personInfo->age;

?>

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;
  }
}

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
}