javascript - 处理 AJAX 调用的成功

Question

我想处理来自下面提交的成功调用并重新加载当前页面。我怎样才能最好地做到这一点？

$('#modal-confirm').click(function() {
    $(form).ajaxSubmit();
    $(form).resetForm();
    $('#myModal').modal('toggle');
});

Answer 1

您可以将选项传递给ajaxSubmit()

$(form).ajaxSubmit({
   success:function(){
     alert("complete");
     // do other stuff
   }
});

Answer 2

你也许可以试试这个..

$('#modal-confirm').on("click", function(e){
  e.preventDefault();
  $.ajax({
    url     : '...'
  }).done(function(data) {
  //handle your conditions here once the request is successful.  if conditions are met, then...
    location.reload(true);
  });
});

Answer 3

    $(document).ajaxSuccess(function() { // http://api.jquery.com/ajaxSuccess/
     location.realod();
    })
    $('#modal-confirm').click(function() {
        $(form).ajaxSubmit();
        $(form).resetForm();
        $('#myModalOne').modal('toggle');
    });