Title rewritten to: Error: "SyntaxError: """ is not a legal JSON format"

Question

I have a question about the following code. The output of console.log is:

The URL I requested via a JavaScript Ajax request was "login.php":

<?php include('init.php');
    use LoginLoginService;

    #include(__DIR__.'/Login/LoginService.php');

    global $pdo;
    session_start();

    $username = $_POST['username'];
    $pass = $_POST['password'];
    if (!empty($username)) {
        $test = new LoginService();
        $user = $test->getUsersLogin($username);
        if (!empty($user) && $user[0]['login'] == $username) {
            $json = json_encode(array("success" => 1));
            echo $json;
        } else {
            $json = json_encode(array("success" => 0));
            echo $json;
        }
    }
    ?>

My JavaScript Ajax request:

$(() => {
    $('.login-form').on('submit', function (e) {
        e.preventDefault();

        $.ajax({
            type: "POST",
            dataType: "json",
            timeout: 500,
            url: '/src/login.php',
            data: $(this).serialize(),

            success: (data) => {
                try {
                    var jso = JSON.parse(data);
                    console.log(jso);
                } catch (e) {
                    console.log(e);
                    return false;
                }
            },
            error: (data) => {
                console.log(JSON.parse(data));
            }
        });
    });
});

Why is PHP's response of {"success":1} incorrect? what is the problem?

SyntaxError: "[object Object]" is not valid JSON

Answer 1

Try this to avoid this error:

myFunction(data: string) {
  try {
    JSON.parse(data); 
    console.log(data);
  }
   catch (e) {
   console.log(e); 
  }
}

Answer 2

If you write dataType: "json", then jQuery will automatically parse your response into JSON before entering the "success" function. This is described in detail in jQuery's $.ajax documentation.

Therefore, data is already an object. You cannot pass an object to JSON.parse() - it expects a string.

So, no need

var jso = JSON.parse(data); console.log(jso);

You can write directly

console.log(data);