C#后台如何接收前端JS發送的formdata對象

function fileUpload() {
    var formData = new FormData();
    formData.append("param1", "參數1");
    formData.append("param2", "參數2");
    formData.append("file", $('#uploadfile')[0].files[0]);
    $.ajax({
        url: 'Dosomething/Doit',
        //url
        type: 'post',
        data: formData,
        processData: false, //必須false才會避開jQuery對 formdata 的默認處理   
        contentType: false, //必須false才會自動加上正確的Content-Type
        async: false
    }).done(function(res) {;
    }).fail(function(res)) {;
    });
}

以上是前端js

[HttpPost]
public bool Doit()
{
    string p1=Request.Form["param1"]; //普通參數獲取
    string p2=Request.Form["param2"];
    var file = Request.Files["file"];
    foreach(string upload in Request.Files.AllKeys)
   {
      HttpPostedFileBase file=Request.Files[uplooad];  //file可能為null
    }

    return true;
}

以上是后端接收.