C#后台如何接收前端JS发送的formdata对象

 1 function fileUpload() {
 2     var formData = new FormData();
 3     formData.append("param1", "参数1");
 4     formData.append("param2", "参数2");
 5     formData.append("file", $('#uploadfile')[0].files[0]);
 6     $.ajax({
 7         url: 'Dosomething/Doit',
 8         //url
 9         type: 'post',
10         data: formData,
11         processData: false, //必须false才会避开jQuery对 formdata 的默认处理   
12         contentType: false, //必须false才会自动加上正确的Content-Type
13         async: false
14     }).done(function(res) {;
15     }).fail(function(res)) {;
16     });
17 }

以上是前端js

 1 [HttpPost]
 2 public bool Doit()
 3 {
 4     string p1=Request.Form["param1"]; //普通参数获取
 5     string p2=Request.Form["param2"];
 6     var file = Request.Files["file"];
 7     foreach(string upload in Request.Files.AllKeys)
 8    {
 9       HttpPostedFileBase file=Request.Files[uplooad];  //file可能为null
10     }
11 
12     return true;
13 }

以上是后端接收.