Ajax运行不正常

Question

I'm trying to submit a form using Ajax , but it doesn't work here is my Ajax :

$(document).ready(function(){
  $("#submit").click(function(event){
var ad1 = $("#ad1").val();
var ad2 = $("ad2").val();
var city = $("city").val();
var state = $("state").val();
var zip = $("zip").val();
var country = $("country").val();
var mm = $("mm").val();
var dd = $("dd").val();
var yy = $("yy").val();
var lname = $("lname").val();


// Returns successful data submission message when the entered information is stored in database.
var dataString = 'name1='+ name + '&ad11='+ ad1 + '&ad21='+ ad2 + '&city1='+ city + '&state1='+ state + '&zip1='+ zip + '&country1='+ country + '&mm1='+ mm + '&yy1='+ yy + '&dd1='+ dd + '&lname1=';

        if(name=='')
        {
            alert("");
        }
        else
        {
            // AJAX Code To Submit Form.
            $.ajax({
                type: "POST",
                url: "action.php",
                data: dataString,
                cache: false,
                success: function(result){
                    alert(result);
                }
            });
        }
        return false;
    });
});

and it's not giving any result just giving data in header

the result is like :

---

I copied the javascript to the form page it's now working ,but the ajax is returning a blank alert while it should be "Form Submitted Succesfully" I guess that it's an error of inclusion of the file , but i'm using the right directories.

---

here is action.php :

<?php
$con = mysqli_connect("server","user","pass","db"); 

$name=$_POST['name1'];
$ad1=$_POST['ad11'];
$ad2=$_POST['ad21'];
$city=$_POST['city1'];
$state=$_POST['state1'];
$zip=$_POST['zip1'];
$country=$_POST['country1'];
$mm=$_POST['mm1'];
$dd=$_POST['dd1'];
$yy=$_POST['yy1'];
$dob=$dd."/".$mm."/".$yy;
$mm=$_POST['mm1'];
$name=$_POST['name1'];
$lname=$_POST['lname1'];


$r2=rand(10000,90000);

$query = mysqli_query($con,"insert into users values('$r2','$name','$lname','$ad1','$ad2','$city','$state','$zip','$country','$dob')");
mysqli_close($con);
echo "Form Submitted Succesfully";

?>

---

Answer 1

This name variable is not have defined in ajax file var dataString = 'name1='+ name + so name would be an empty string

=> if(name=='') { alert(""); } executed. Please add string into alert and check again :)