jQuery Ajax函数不起作用？

I'm developing a simple login form for my website. And to do that I thought to use ajax to connect with php to validate users. However to do that I cannot get output from ajax.

<script>
 function submitForLogin() 
 {  
    $.ajax({
            type: "POST",
            url: "php/login.php",
            data: { email: "example@abc.com",password:"123" }}).done(function(data){alert(data);});

 }

</script>

When user clicks on login button it calls submitForLogin() function.

Above part of the code I've placed in my login.html file. To validate whether this works or not I simply replaced data values of email and password with hard coded values.

Note : example@abc.com and 123 both email and password stored in Wamp server database.

This is the PHP file :

<?php
$userEmail=$_POST['email'];
$userPass=$_POST['password'];
$servername ="localhost";
    $username="root";
    $password="";
    $dbname="AS2014459";
    //To create a connection
    $con = mysqli_connect($servername,$username,$password,$dbname); //check connection
    if(!$con){
        die("Connection failed: ".mysqli_connect_error());
    }   
    $sql="SELECT Email,Password FROM USERTABLE WHERE Email='".$userEmail."'";
    $results=mysqli_query($con,$sql);
    if(mysqli_num_rows($results)>0)
    {
        echo "userExist";
    }
    else
    {
        echo "fakeUser";
    }
    mysqli_close($con);



?>

Whenever I run php file only (with $userEmail and $userPass having previous hard coded values) php prints userExist output. But using ajax I cannot get that in an alert box.

Is there something I missing? I'm running the website in wamp server too.

UPDATE

When I check console errors it shows;

enter image description here

And when I click on login.html line 112, it shows;

enter image description here

And ideas guys? Also, none of the solutions provided so far gave me successful answer for the question.

There was a jquery error and now it's fixed. But syntax error exists.

Try this code working for me on my machine. Don't forgot to place Jquery File. PHP (ajax.php) :

<html>
<head>
    <script src="jquery.min.js"></script>
    <script>
        function submitForLogin() 
 {
        alert("23");
        var email = document.getElementById("txtUserName").value;
        var password = document.getElementById("txtPassword").value;
    $.ajax({
            type: "POST",
            url: "login1.php",
            data: { email: email,password:password },
            success : function(response){
                alert(response);
            }
   });

 }

    </script>
</head>
<body>
    <input type="text" placeholder = "Enter user name" id="txtUserName"/>
    <input type="password" placeholder = "Enter password" id="txtPassword"/><br>
    <input type="button" onclick="submitForLogin();" value="Login"/>
</body>
</html>

Server Side (login1.php) :

    <?php
var_dump($_POST);
?>

You need to include a function for when the request is successful, which includes the response.

$.ajax({
    type: "POST",
    url: "php/login.php",
    data: { email: "example@abc.com",password:"123" },
    success:function(data){
        alert(data)
    };
});

$.ajax({
    type: "POST",
    url: "php/login.php",
     data: { email: "example@abc.com",password:"123" },
    success: function (data) {
        alert(data);

    }
});

your data in AJAX call is in JSON format. in your php script you should use json decode as below

$data = file_get_contents("php://input");
$data = json_decode($data,true);
$userEmail=$data['email'];
$userPass=$data['password'];

i hope this might help you. your login will work as u expect it

Change your php file like below :

<?php
$userEmail=$_POST['email'];
$userPass=$_POST['password'];
$servername ="localhost";
$username="root";
$password="";
$dbname="AS2014459";
//To create a connection

header('Content-type: application/json');


$con = mysqli_connect($servername,$username,$password,$dbname); //check connection
if(!$con){
    die("Connection failed: ".mysqli_connect_error());
}   
$sql="SELECT Email,Password FROM USERTABLE WHERE Email='".$userEmail."'";
$results=mysqli_query($con,$sql);
if(mysqli_num_rows($results)>0)
{
     $res = "userExist";
     $data = json_encode($res);
     echo $data;
     die();

}
else
{

    $res = "fakeUser";
    $data = json_encode($res);
    echo $data;
    die();
}
mysqli_close($con);



?>

And your script :

$.ajax({
type: "POST",
url: "php/login.php",
data: { email: "example@abc.com",password:"123" },
success:function(data){
    alert(data);
};
});