Ajax 将参数传递给onSuccess回调函数

Ajax 将参数传递给onSuccess回调函数,ajax,asp.net-mvc,forms,Ajax,Asp.net Mvc,Forms,我有以下表格: @using (Ajax.BeginForm("DoComment", "Publication", new { id = Model.Publication.OID, parentId = Model.OID }, new AjaxOptions { OnSuccess = "publishCommentSuccess" })) { //some code } 如何将par

我有以下表格:

   @using (Ajax.BeginForm("DoComment", "Publication", new { id = Model.Publication.OID, parentId = Model.OID }, new AjaxOptions
         {
              OnSuccess = "publishCommentSuccess"
         }))
   {
           //some code
   }
如何将
parentId
传递到
publishCommentSuccess
函数中

function publishCommentSuccess (json) {

        if (!json.Result)
            showAuthBox();
        else
            alert(json.Message);

    };  

谢谢。

由于此参数已传递给控制器操作,您可以让此控制器操作将其作为JSON响应的一部分返回:

public ActionResult DoComment(int id, int parentId)
{
    ... 

    return Json(new 
    {
        Message = "foo",
        ParentId = parentId
    }, JsonRequestBehavior.AllowGet);
}
以便您可以在成功回调中使用它:

function publishCommentSuccess (json) {
    alert(json.ParentId);

    ...
};