How to pass model from one partial view to another partial view - asp.net

Question

I have a model called Result. Suppose this has 4 fields, like student_id, marks, status, remarks.
Now I have a view in which student listing is shown. In front of each student, there is a button for enter marks of exam. On clicking on button a pop-up will open and there will be 2 fields student_id, marks and 2 buttons 'pass' and 'fail'.
On clicking on fail button another pop-up will appear for enter remarks only.
Now my question is that, how can I retain values of first pop-up on second pop-up, As on clicking on 'submit' button of second pop-up, I will save all the details.
I know a way to do this using hidden fields in second pop-up. Is there any other way to do this?
Model classes are:
1. User (id, name, f_name, address...)
2. Result (student_id, marks, grade, remarks)
Student List view
#{
List<User> Student = (List<User>)ViewData["Student"];
}
<table id="table_id">
<tr>
<th class="dbtc">S.No.</th>
<th class="dbtc">Student Name)</th>
<th style="width: 110px">Operate</th>
</tr>
#foreach (User usr in Student)
{
int index = Student.IndexOf(usr);
<tr>
<td class="dbtc">
#(Student.ToList().IndexOf(usr) + 1)
</td>
<td>
#Html.ActionLink(usr.FirstName + " " + usr.LastName, "Details", "User", new { id = usr.Id }, null)
</td>
<td>
#Ajax.ActionLink("Examine", "Result", new { id = Model.Id, userId = usr.Id }, new AjaxOptions
{
HttpMethod = "GET",
UpdateTargetId = "divPopup",
InsertionMode = InsertionMode.Replace,
OnSuccess = "openPopup('Examine Content')"
})
</td>
</tr>
First Partial view of examine
#model ComiValve.Models.Result
#using (Html.BeginForm("ExamPass", "Student", new { #id = (int)ViewBag.id, userId = (int)ViewData["UserId"] }, FormMethod.Post))
{
<div id="divExamAdvice"></div>
<div class="editor-label">
#Html.DisplayNameFor(model => model.Name)
</div>
<div class="editor-field">
#Html.TextAreaFor(model => model.Marks)
</div>
<div class="editor-field">
#Html.TextAreaFor(model => model.Grade)
</div>
<div class="login_submit_div">
<input type="submit" value="Pass" />
#Ajax.ActionLink("Fail", "ExamAdvice", new { id = (int)ViewBag.id, userId = (int)ViewData["UserId"] }, new AjaxOptions
{
HttpMethod = "GET",
UpdateTargetId = "divPopup",
OnSuccess = "openPopup('Exam Advice')"
})
</div>
}
Second partial view for remaks (when user click on fail, then this view will open.)
#model ComiValve.Models.ExamContent
#using (Html.BeginForm("ExamFail", "Student", new { id = Model.id }, FormMethod.Post))
{
<div id="divExamAdvice"></div>
<div class="editor-label">
#Html.DisplayNameFor(model => model.Remarks)
</div>
<div class="editor-field">
#Html.TextAreaFor(model => model.Remarks)
</div>
<div class="left">
<input type="submit" value="Confirm Fail" />
</div>
}
Methods of Controller
public virtual ActionResult ExamContent(int id, int userId)
{
ViewBag.IsApprove = true;
ViewBag.UserId = userId;
ViewBag.id = id;
return PartialView("ExamContent");
}
public virtual ActionResult ExamAdvice(int id, int userId)
{
ViewBag.IsApprove = true;
if (Request.IsAjaxRequest())
{
Result result = new Result();
result.id = id;
result.User = db.Users.Find(userId);
return PartialView("ExamAdvice", result);
}
else
{
return RedirectToAction("Index");
}
}

Answer 1

Why are you passing the model between partial views. You can create a single Model and use it on both the views. In case of having two different tables, create the two different "Lists" of "Table" type. Like this
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.ComponentModel.DataAnnotations;
using System.ComponentModel;
using System.Web.Mvc;
namespace LearningMVCapp.Models
{
public class Data
{
public List<tbl_Dept> lstDepatrment;
public List<tbl_employees> lstEmployees;
//other properties
}
}
You can also use session instead of hidden fields, refer this link http://www.dotnet-tricks.com/Tutorial/mvc/906b060113-Controlling-Session-Behavior-in-Asp.Net-MVC4.html.