如何使用此代码无法命中我的 ASP.NET MVC 应用程序中的控制器

Question

<asp:Content ID="Content2" ContentPlaceHolderID="MainContent" runat="server">
<script type="text/javascript">
    GOTO = function () {
        alert("yes");
        $.ajax({
            cache: false,
            type: "POST",
            url: "/Home/Index/",
            data: datastring,
            dataType: "json",
            success: function (data) {
                alert("Ohh Yaa Success");
            }
        });
    }
</script>

   <input type="button" value="submit" onclick="JavaScript:GOTO()" />

</asp:Content>

我的控制器 ActionResult 是这样的 JsonResult

  [HttpPost]
        public System.Web.Mvc.JsonResult Index(FormCollection collection)
        {
            //return Content("<xml>this is just test</xml>", "text/xml");
            //return Content("this is just test", "text/plain");

            if (Request.AcceptTypes.Contains("application/json"))
            {
                return Json(new { id = 1, value = "new" });
            }
            else if (Request.AcceptTypes.Contains("application/xml") ||
                     Request.AcceptTypes.Contains("text/xml"))
            {

            }
            if (Request.AcceptTypes.Contains("text/html"))
            {
                //return View();
            }

           return Json(new { foo = "bar", baz = "Blech" });
        }

我始终无法在此处返回 JsonResult 我收到弹出消息，说您已选择打开此对话框？我做错了什么吗？

谢谢

Answer 1

试试这个——并确保首先加载 jQuery。请注意通过 jQuery 应用处理程序（而不是内联）、序列化数据、在代码中动态生成 URL（而不是硬编码）以及从单击处理程序返回 false 以阻止正常表单提交的更改。

<script type="text/javascript">
    $(function() {
        $('input[type=button]').click( function() {
          var data = $('form').serialize();  // or however you get your data
          $.ajax({
              cache: false,
              type: "POST",
              url: "<%= Html.Action( "index", "home" ) %>",
              data: data,
              dataType: "json",
              success: function (data) {
                  alert("Ohh Yaa Success");
              }
          });
          return false; // don't do the normal submit
        });
    });
</script>

<input type="button" value="submit" />

Answer 2

您需要将按钮放在表单标签中并在 onsubmit 事件中调用 GOTO 函数

Answer 3

看来您的 data: datastring 可能是问题所在。检查并确保数据参数的名称与方法参数的名称相同。

Answer 4

我会尝试更像这样......

<asp:Content ID="Content2" ContentPlaceHolderID="MainContent" runat="server">
<script type="text/javascript">
    $(document).ready(function () {
        $(form).submit(function() {
        alert("yes");
        $.post({
            cache: false,
            type: "POST",
            url: "/Home/Index/",
            data: datastring,
            dataType: "json",
            success: function (data) {
                alert("Ohh Yaa Success");
            }
        });
      });
    }
</script>

<form>
   // your form fields
   <input type="button" value="submit" />
</form>
</asp:Content>

然后你的控制器应该看起来更像这样。

请注意我们如何将参数更改为与您的 jQuery data 字段匹配的字符串。

[HttpPost]
public System.Web.Mvc.JsonResult Index(string datastring)
{
    // you can deserialize your Json here.

    //return Content("<xml>this is just test</xml>", "text/xml");
    //return Content("this is just test", "text/plain");

    if (Request.AcceptTypes.Contains("application/json"))
    {
        return Json(new { id = 1, value = "new" });
    }
    else if (Request.AcceptTypes.Contains("application/xml") ||
             Request.AcceptTypes.Contains("text/xml"))
    {

    }
    if (Request.AcceptTypes.Contains("text/html"))
    {
        //return View();
    }

   return Json(new { foo = "bar", baz = "Blech" });
}