Php try json decode

Php json_decode

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

$person = json_decode($personJSON);

echo $person->name; // Johny Carson

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
}

-1

Similar codes

3 votes282 views4 answers
How to get last 10 digit from number in php
PHP
0 vote313 views1 answer
Codeigniter sms send
PHP
0 vote335 views1 answer
Add code return block phpstorm
PHP
0 vote322 views1 answer
Remove nul value aray php
PHP
1 vote308 views1 answer
Command "make:controller" is not defined.
PHP
2 votes286 views2 answers
How to use vue 3 in laravel 8
PHP
0 vote291 views1 answer
Access to this resource on the server is denied laravel
PHP
0 vote361 views1 answer
Laravel get file to browser send
PHP Laravel
0 vote286 views4 answers
Mysqli_query() expects parameter 1 to be mysqli, string given
PHP
4 votes299 views1 answer
Php where is ini file
PHP