阿贾克斯没有回应

I try to take a response but I could not. I can't describe it why it does not work or what it is wrong.

var virmanYap = function(){
   $("#loading").show();
   $("#tblVirman").hide();
   alert('Virman');
   var data = $("#virman_filtre").serialize();
   $.post("php/virman_yap.php", data).success(function(r){
        alert(r);
   });
}

my php code:

foreach ($_POST['virman'] as $evrakNo => $detay) {
    print_r($_detay);
}
echo "asd";

Try $.ajax instead of $.post:

var virmanYap = function () {
    $("#loading").show();
    $("#tblVirman").hide();
    alert('Virman');
    var data = $("#virman_filtre").serialize();
    $.ajax({
        url: 'php/virman_yap.php',
        type: 'POST',
        data: data,
        success: function (r) {
            alert(r);
        }
    });
}

Check $.post documentation here: https://api.jquery.com/jQuery.post/

As you can see, success callback is one of the parameters, you don't need to chain it.

$.post("php/virman_yap.php", data, function(r){
        alert(r);
});

Do something like this. Just remove the chained success function and you should be good.