php ajax返回假问题

Question

I am using the following ajax function

function validatec()
{
    var coupon = jQuery('#validateC').val();
    var data = "coupon="+coupon;

        $.ajax({
            type      : 'POST', //Method type
            url       : 'https://sainicomputers.co.in/exam/index.php/login/test/', //Your form processing file URL
            data      : data, 
            async : false, 
            success: function(msg) {
                if(msg > 0)
                {

                }
                else
                {
                    alert("Please enter a valid Coupon Code");

                }
                 return false; 

            }

        });
}

What I want to do is that when the result is received from ajax call then I dont want to load the page. but this code is not working how can I do this ?

Answer 1

Your code not showing that there is a cause to page refresh, i think that you are calling the ajax when you click o button of type 'submit', if yes, you have to change the type from submit to button and in your success reponse you have to add somthing like:

jQuery('#form').submit();