AJAX data.Success在响应中有数据时返回false

Question

我正在使用JQuery Form Plugin,但是IE 11存在问题,其中"data.Success"会返回false,尽管响应中有JSON,可以通过使用IE Developer Tools跟踪网络流量来证明.

我看到一个成功的初始AJAX POST到/ NewClubOpeningTool/ProcessCompose /带有水合JSON响应,但data.Success是错误的.

响应:

{"Success":true,"ViewMode":"Prospects","DraftOrSentViewMode":"Draft","NewClubId":21}

适用于FF和Chrome.

任何想法,为什么会发生这种情况以及我可以尝试什么？

//表单设置

@using (Ajax.BeginForm("ProcessCompose", "NewClubOpeningTool", FormMethod.Post, new AjaxOptions { UpdateTargetId = "update_panel" }, new { @enctype = "multipart/form-data", id="ComposeForm" }))   

//处理表单提交的控制器操作

       [HttpPost]
        public ActionResult ProcessCompose(NewClubInviteComposeEmailViewModel compose, IEnumerable<int> Recipients)
        {
            [...]

            return Json(new { Success = success }, "text/json", System.Text.Encoding.UTF8, JsonRequestBehavior.AllowGet);
        }




 //This code lives in the view that has the AJAX form
 $(document).ready(function () {

        /***********************************************************
            Initialize the JQuery Form object
            This is how we are able to upload stuff via an AJAX form
        *************************************************************/

        var options = { 
            type:'post',
            data: $('#ComposeForm').serializeArray(),
            datatype: 'json',
            url: '/NewClubOpeningTool/ProcessCompose/',
            success: function (data) {
            if (data.Success) {
                alert('Success! We have data.Success!');
                $(".dialog").remove();
                $("#divInviteContentArea").load("/NewClubOpeningTool/InviteEmail?clubId=" + data.NewClubId + "&viewMode=" + data.ViewMode + "&draftOrSentViewMode=" + data.DraftOrSentViewMode);
            } else {
                alert('Error! We do not have data.Success!');
            }
        },
        error: function(jqXHR, textStatus) {
            alert('Error: ' + textStatus);
        }                   
    }; 
    $('#ComposeForm').ajaxForm(options);
        /*************************************************************
        Prevent the AJAX form from submitting twice
        **************************************************************/
        $("#ComposeForm").submit(function (e) {

             e.preventDefault();

             return false;


        });
    });

Answer 1

一切都是正确的,但你犯了一个小错误:即: -

改变datatype:'json'到dataType:'json'其Ť在AJAX调用(CAPS).