Question

我正在开发MVC网站。我想逐个显示详细记录。用户单击“提交”按钮时显示的下一条记录。请提供相同的建议控制器中的代码

  // GET: Member_Details
   public ActionResult Member_Details(int sysmemberid , string type)
        {
            Member_Details obj = new Member_Details();
            obj=obj.getMemberDetails(Convert.ToInt64(Session["Id"]), sysmemberid, type);// assign values to model
            Session["sysmemberid"] = sysmemberid;
            Session["type"] = type;
            Session["Next"] = obj.next;
            return View(obj);
        }



 [HttpPost]
     public void Member_Details(Member_Details obj ,string command)
     {
         string sysmemberid = Session["sysmemberid"].ToString();
         string type1 = Session["type"].ToString();
         string sqlQuery = "";
         int next = Convert.ToInt32(Session["Next"]);   
         RedirectToAction("Member_Details", new { sysmemberid = next, type = type1 });
}

查看代码：

@using (Html.BeginForm("Member_Details", "Member_Details", FormMethod.Post, new { id = "submitForm" }))
{
   // Code to bind model
   <button type="submit" id="btnSave" name="command" value="invite" class="btn btn-lg btn-primary btn-block link dtlSubmit">Connect</button>

}

Answer 1

看起来你已经有了获得下一个项目并构建ReditectToAction的逻辑，但它看起来并不像你要回来的那样。

将ActionResult返回值添加到处理POST的Member_Details：

public ActionResult Member_Details(Member_Details obj ,string command)

并返回RedirectToAction：

     return RedirectToAction("Member_Details", new { sysmemberid = next, type = type1 });

将多个传递值传递给MVC 5中的RedirectToAction

1 个答案: