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CODE EXAMPLE FOR PHP

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;
  }
}
Source by stackoverflow.com #
 
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Tagged: #php #json #decode
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