使用AJAX提交表单并根据表单中的数据跳转到特定页面

Question

我有一个带有搜索框的表单。根据用户搜索的内容，我会跳转到特定页面。使用$ .ajax如何跳转到该URL并相应地将数据发送到该URL？

HTML：

<form method = "POST">
<input id = "search"...>
</form>

JS

//When user presses enter submit the form, jump to the url


$("#search").keyup(function(event){
    if(event.keyCode == 13){
        form.submit();
    }

            var landingUrl = '/page/subpage/'; //page changes depending on what the user searched for so I can't put <form action = "/page/subpage/">
            var data = '...';

            $.ajax({
              type: "POST",
              data: {
                  data
              },
              url: (landingUrl),
              contentType: "application/x-www-form-urlencoded",
              success: function() {
                  console.log("post data sent");
              },
              error: function() {
                  $window.open(landingUrl, "_self");
              }
          });
         form.submit();
    });

如果您需要更多信息，请不要犹豫。