jQuery AJAX Script返回parseerror

Question

我有一个我正在构建的注册表单，应该处理表单提交的AJAX代码会一直返回parseerror。 我正在使用Material Design Framework。

这是jQuery代码：

//表单提交

$form.submit(function () {
    $('html, body').animate({
        scrollTop: 0
    }, 800);

    $.ajax({
        type: "POST",
        url: '/libraries/signup.php',            
        dataType: "text",
        data: {
            firstName: $fName.val(),
            otherNames: $oName.val(),
            userEmail: $uEmail.val(),
            phoneNumber: $pNumber.val(),
            userPassword: $uPassword.val(),
            userGender: $gender.val(),
            accType: $accType.val(),
            submit: 'Sign Up'
        },
        beforeSend: function () {
            $("input[type=submit]").prop("disabled", true);
            $("#p2").fadeIn(120);
            $("div.error").fadeOut(380).empty();
            showToast("Please Wait...");
        },
        success: function (response) {
            $("input[type=submit]").prop("disabled", false);
            if (response.type === "success") {
                showToast(response.content);
            } else {
                $("div.error").slideUp().html(response.content).fadeIn(500);
                setTimeout('$("div.error").fadeOut(500);', 3500);
                $("#p2").fadeOut(320);
            }
        },
        error: function (jXHR, textStatus) {
            $("input[type=submit]").prop("disabled", false);

            $("div.error").slideUp().html("Couldn't submit request: " + textStatus).fadeIn(500);
            setTimeout('$("div.error").fadeOut(500);', 5500);
            $("#p2").fadeOut(320);
        }
    });
    return false;
});

这是PHP：

//Configurations File
include dirname(__FILE__)."/functions.php";

//Check Form Submission
if($_SERVER['REQUEST_METHOD'] == "POST" && $_POST['submit'] == "Sign Up"){

    //Sanitize and Encrypt User Input
    $firstName = filterUserInput($_POST['firstName']);
    $otherNames = filterUserInput($_POST['otherNames']);
    $userEmail = filterUserInput($_POST['userEmail']);
    $phoneNumber = filterUserInput($_POST['phoneNumber']);
    $userPassword = filterUserInput($_POST['userPassword']);
    $gender = filterUserInput($_POST['userGender']);
    $accountType = filterUserInput($_POST['accType']);  

    registerMember($firstName, $otherNames, $userEmail, $phoneNumber, $userPassword, $gender, $accountType, $db);
}

我尝试了几种方法并在线查看了这么多解决方案，但问题仍然存在。 任何帮助都非常感谢。