Submit form via Ajax

Question

So, I followed a tutorial here to be able to submit a form with ajax. I followed the tutorial exactly (atleast I thought I did) and when I try to submit the form the page just refreshes and it never gets to the php script to send it to the database.

The script that I am using is below:

$(function () {
$(".button").click(function () {
    $(function () {
        $('.error').hide();
        $("#submit_btn").click(function () {
            //validate and process form here

            $('.error').hide();
            var firstname = $("input#firstname").val();
            if (firstname == "") {
                $("label#firstname_error").show();
                $("input#firstname").focus();
                return false;
            }

            var lastname = $("input#lastname").val();
            if (lastname == "") {
                $("label#lastname_error").show();
                $("input#lastname").focus();
                return false;
            }

            var email = $("input#email").val();
            if (email == "") {
                $("label#email_error").show();
                $("input#email").focus();
                return false;
            }

            var pin = $("input#parent_pin").val();
            if (pin == "") {
                $("label#parent_pin_error").show();
                $("input#parent_pin").focus();
                return false;
            }

            var login = $("input#login").val();
            if (login == "") {
                $("label#login_error").show();
                $("input#login").focus();
                return false;
            }

            var passwd = $("input#passwd").val();
            if (passwd == "") {
                $("label#passwd_error").show();
                $("input#passwd").focus();
                return false;
            }

            var cpasswd = $("input#cpasswd").val();
            if (cpasswd == "") {
                $("label#cpasswd_error").show();
                $("input#cpasswd").focus();
                return false;
            }

            var user_type = $("input#user_type").val();
            if (user_type == "") {
                $("label#user_type_error").show();
                $("input#user_type").focus();
                return false;
            }

            var dataString = 'firstname=' + firstname + '&lastname=' + lastname + '&email=' + email + '&parent_pin=' + pin + '&login='
            login + '&passwd='
            passwd + 'user_type' = user_type;
            //alert (dataString);return false;
            $.ajax({
                type: "POST",
                url: "studentAccess/files/AddNewUser.php",
                data: dataString,
                success: function () {
                    $('#form-body').html("<div id='message'></div>");
                    $('#message').html("<h2>New User Added Successfully!</h2>");

                }
            });
        });
    });
  });
});

The error that I am receiving in Google Chrome's console is:

Uncaught SyntaxError: Unexpected identifier AddNewUser.js:65

Line 65 would be:

var dataString = 'firstname=' + firstname + '&lastname=' + lastname + '&email=' + email + '&parent_pin=' + pin + '&login=' login + '&passwd=' passwd + 'user_type' = user_type;

I'm not sure how to fix this problem because I don't know what the error means. Any help would be great!

UPDATE

<?php
$con = mysqli_connect("");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql = "INSERT INTO members (firstname, lastname, email, login, psswd, user_type)
VALUES
('$_POST[firstname]','$_POST[lastname]','$_POST[email]', '$_POST[login]', '$_POST[psswd]', '$_POST[user_type]')";

if (!mysqli_query($con, $sql)) {
    die('Error: ' . mysqli_error($con));
}
echo "1 record added";

mysqli_close($con);
?>

Answer 1

In the below line, it should be '&user_type=' + user_type; instead of 'user_type' = user_type;

var dataString = 'firstname=' + firstname + '&lastname=' + lastname + '&email=' + email + '&parent_pin=' + pin + '&login=' login + '&passwd=' passwd + 'user_type' = user_type;

Also, if the button is a submit button, you should prevent the default form submit action using event.preventDefault(); within the button's click event.

In addition, the first two lines of code is not required. I have commented them out.

/*$(function () {
$(".button").click(function () {*/

    $(function () {
        $('.error').hide();
        $("#submit_btn").click(function (event) { 
            event.preventDefault();
            //rest of your current validation code should be put here.
        });
     });

Answer 2

At the very end of that line you have:

'user_type' = user_type;

It needs to be:

'&user_type=' + user_type;

You may also need to add return false; after your ajax to prevent the page from refreshing (and clearing out your form).

$.ajax({
    type: "POST",
    url: "studentAccess/files/AddNewUser.php",
    data: dataString,
    success: function () {
        $('#form-body').html("<div id='message'></div>");
        $('#message').html("<h2>New User Added Successfully!</h2>");
    }
});
return false; //Keep page from refreshing

Further EDIT:

You also have a .click() embedded in a .click(). You cannot click two buttons at one time. You need to change this:

$(function () {
$(".button").click(function () {
    $(function () {
        $('.error').hide();
        $("#submit_btn").click(function () {

to this...

$(function () {
    $('.error').hide();
    $("#submit_btn").click(function (e) {
        e.preventDefault();
        ...

Answer 3

You have missed + symbols in the #65 line

Should be

var dataString = 'firstname=' + firstname + '&lastname=' + lastname + '&email=' + email + '&parent_pin=' + pin + '&login=' + login + '&passwd=' + passwd + '&user_type=' + user_type;

Please read http://en.wikipedia.org/wiki/JavaScript_syntax

Remove $(function () { in .button click handler. Now it just registers a handle on button click but not executes it