Formdan gelen bilgileri ajaxla WordPress'teki bir sayfama post etmek istiyorum. Formda bulunan butonu tıkladığımda saveform() işlemi yapıyor.
Butona tıklandığında https://websitem.com/onayla/ sayfasına gitsin ve post edilen bilgileri göstersin istiyorum.
Ajax konusunda bilgi sahibi arkadaşların yardımına ihtiyacım var.



function saveform(){
var datastring = $(".form-wrapper").serialize();
datastring = datastring + "&price=" + formatMoney(price) + " TL";
datastring = datastring + "&km=" + formatMoney(km);
datastring = datastring + "&date=" + $(".dates-bar").find(".selected").data("date");

datastring = datastring + "&CityList1Name=" + $('.CityList1 option:selected').text();
datastring = datastring + "&DistrictList1Name=" + $('.DistrictList1 option:selected').text();
datastring = datastring + "&NeighborhoodList1Name=" + $('.NeighborhoodList1 option:selected').text();

datastring = datastring + "&CityList2Name=" + $('.CityList2 option:selected').text();
datastring = datastring + "&DistrictList2Name=" + $('.DistrictList2 option:selected').text();
datastring = datastring + "&NeighborhoodList2Name=" + $('.NeighborhoodList2 option:selected').text();

datastring = datastring + "&price2text=" + $("[for=" + $("input[name=price2]:checked").attr("id") + "]").text();
datastring = datastring + "&price3text=" + $("[for=" + $("input[name=price3]:checked").attr("id") + "]").text();
datastring = datastring + "&price4text=" + $("[for=" + $("input[name=price4]:checked").attr("id") + "]").text();
datastring = datastring + "&price5text=" + $("[for=" + $("input[name=price5]:checked").attr("id") + "]").text();
datastring = datastring + "&price6text=" + $("[for=" + $("input[name=price6]:checked").attr("id") + "]").text();
datastring = datastring + "&price7text=" + $("[for=" + $("input[name=price7]:checked").attr("id") + "]").text();
datastring = datastring + "&price8text=" + $("[for=" + $("input[name=price8]:checked").attr("id") + "]").text();
datastring = datastring + "&boxtext=" + $("[for=" + $("input[name=box]:checked").attr("id") + "]").text();


$.ajax({
type: "POST",
url: 'https://websitem.com/onayla/',
data: datastring,
dataType: "json",
success: function(data) {


});
}