Ajax.BeginForm（），OnSuccess - 獲取事件目標

Question

我無法獲得在Ajax.BeginForm()觸發OnSuccess()方法的元素的目標。 所以這是代碼片段：

@using (Ajax.BeginForm("InsertModel", "Home", new AjaxOptions
{
    HttpMethod = "POST",
    OnSuccess = "doWork(this,'SomeCustomText')"
}))
{
    <div id="container">
       <--Some HTML--!>
      <input type="submit" value="OK"/>
    </div>
}

<script>
function doWork(e,customText)
{
   alert(customText);  //It shows 'SomeCustomText', so its good.

   alert(e);  //[Object] object
   alert(e.prop("tagName"));  //Object #<Object> has no method 'prop' 
   alert(e.attr("tagName"));  //Object #<Object> has no method 'attr' 
   alert(jQuery(e).html());  //undefined
   alert(jQuery(e).prop("tagName"));  //undefined
   alert(e.target);  //undefined
   alert(jQuery(e).target);  //undefined
 }
<script/>

問題：

如何獲得目標？！ 謝謝

更新1

jQuery版本應該如下所示：

jQuery.ajax({
    url:"/Home/InsertModel",
    data:"some post data",
    type:"POST",
    success:function(data){ 
          doWork(this,data); // so i really do not care about data! I just need jQuery(this)
    }
  })

Answer 1

如果要訪問表單元素，則有很多選項，如果頁面上只有一個表單，則可以執行$("form") ，jquery將為您提供表單元素。

另一個選擇是更改Ajax.BeginForm構造函數，它將表單id作為參數，如下所示：

@using (Ajax.BeginForm("InsertModel", "Home",null, new AjaxOptions
{
    HttpMethod = "POST",
    OnSuccess = "doWork('SomeCustomText')"
}, new {id = "myFormId"}))
{
    <div id="container">
       <--Some HTML--!>
      <input type="submit" value="OK"/>
    </div>
}

並在JavaScript中

<script>
function doWork(customText)
{
   alert(customText);  //It shows 'SomeCustomText', so its good.

   // find the form by id
   $("#myForm1");

   // find forms in the page
   $("form")


 }
<script/>

在簡單的JQuery中：

$.ajax({
    url:"/Home/InsertModel",
    data:"some post data",
    type:"POST",
    success:function(data){ 
       // find the form by id
       $("#myForm1");

       // find forms in the page
       $("form")        

       ...
    }
  });