Fastest way to check if a string is JSON in PHP?

function isJson($string) {   json_decode($string);   return json_last_error() === JSON_ERROR_NONE;}

Answer to the Question

The function json_last_error returns the last error occurred during the JSON encoding and decoding. So the fastest way to check the valid JSON is

// decode the JSON data// set second parameter boolean TRUE for associative array output.$result = json_decode($json);if (json_last_error() === JSON_ERROR_NONE) {    // JSON is valid}// OR this is equivalentif (json_last_error() === 0) {    // JSON is valid}

Note that json_last_error is supported in PHP >= 5.3.0 only.

Full program to check the exact ERROR

It is always good to know the exact error during the development time. Here is full program to check the exact error based on PHP docs.

function json_validate($string){    // decode the JSON data    $result = json_decode($string);    // switch and check possible JSON errors    switch (json_last_error()) {        case JSON_ERROR_NONE:            $error = ''; // JSON is valid // No error has occurred            break;        case JSON_ERROR_DEPTH:            $error = 'The maximum stack depth has been exceeded.';            break;        case JSON_ERROR_STATE_MISMATCH:            $error = 'Invalid or malformed JSON.';            break;        case JSON_ERROR_CTRL_CHAR:            $error = 'Control character error, possibly incorrectly encoded.';            break;        case JSON_ERROR_SYNTAX:            $error = 'Syntax error, malformed JSON.';            break;        // PHP >= 5.3.3        case JSON_ERROR_UTF8:            $error = 'Malformed UTF-8 characters, possibly incorrectly encoded.';            break;        // PHP >= 5.5.0        case JSON_ERROR_RECURSION:            $error = 'One or more recursive references in the value to be encoded.';            break;        // PHP >= 5.5.0        case JSON_ERROR_INF_OR_NAN:            $error = 'One or more NAN or INF values in the value to be encoded.';            break;        case JSON_ERROR_UNSUPPORTED_TYPE:            $error = 'A value of a type that cannot be encoded was given.';            break;        default:            $error = 'Unknown JSON error occured.';            break;    }    if ($error !== '') {        // throw the Exception or exit // or whatever :)        exit($error);    }    // everything is OK    return $result;}

Testing with Valid JSON INPUT

$json = '[{"user_id":13,"username":"stack"},{"user_id":14,"username":"over"}]';$output = json_validate($json);print_r($output);

Valid OUTPUT

Array(    [0] => stdClass Object        (            [user_id] => 13            [username] => stack        )    [1] => stdClass Object        (            [user_id] => 14            [username] => over        ))

Testing with invalid JSON

$json = '{background-color:yellow;color:#000;padding:10px;width:650px;}';$output = json_validate($json);print_r($output);

Invalid OUTPUT

Syntax error, malformed JSON.

Extra note for (PHP >= 5.2 && PHP < 5.3.0)

Since json_last_error is not supported in PHP 5.2, you can check if the encoding or decoding returns boolean FALSE. Here is an example

// decode the JSON data$result = json_decode($json);if ($result === FALSE) {    // JSON is invalid}

Hope this is helpful. Happy Coding!

All you really need to do is this...

if (is_object(json_decode($MyJSONArray))) {     ... do something ...}

This request does not require a separate function even. Just wrap is_object around json_decode and move on. Seems this solution has people putting way too much thought into it.