ajax发布请求到没有响应的PHP（每次都是假设）

Question

I'm using jquery multiselect plugin and I want to perform an ajax request on a select/deselect event.

My problem: When I send the request to the php file, ISSET(VAR) returns every time false so that I can't pass variable to the php file.

But Firebug extension for Chrome/Firefox shows me that the POST value is set right POST -> "Response myvar" but GET is empty.

How do I pass the variable to the php file? (I've searched arround the web but found nothing.)

My script, where this pointer is from the multiselect plugin and afterSelect returns if a object is selected

afterSelect: function()
    {
        this.qs1.cache();
        this.qs2.cache();
        count++;     

        var dataString = "count=" + count;

        if ( count > 0 )
        {
            $.ajax
            ({
                type: 'POST',
                url: 'page-to-send-request.php',
                data: dataString,
                success: function()
                {
                    $("#div-to-load").load("page.php #div-to-load").fadeIn('slow');
                },
                error: function (xhr, ajaxOptions, thrownError)
                {
                    alert(xhr.status);
                    alert(thrownError);
                }
             });
        }
    },

The php page to load has for test only

if($_POST['count'])
{
  $count = $_POST['count'];
  echo "Count " .$count;
}
else{ echo "FALSE"; }

Expected result should be

Count 5

But real output is

FALSE

Answer 1

1st try

success: function(data)
       {
         console.log(data);
       },

it should output "Count " .$count; if not

try to use

var dataString = {count: count};

and then you can use

$("#div-to-load").html(data).fadeIn('slow');

the reason of return false .. you used the .load to load the php page .. when you load it .. javascript know it as a separated page .. so $_POST['count'] is undefined in this case . and It will return False always

your success function should be like this

success: function(data)
           {
              $("#div-to-load").html(data).fadeIn('slow');
           },