嗨,人们遇到问题时收到以下错误:
ObjectContext 实例已被释放,不能再用于需要连接的操作。
并且错误针对以下内容:
@Html.DisplayFor(modelItem => item.tblConsole.ConsoleName) "within the game table"
我的游戏控制器中有以下代码:
using System;
using System.Collections.Generic;
using System.Data;
using System.Data.Entity;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using GameTest.Models;
namespace GameTest.Controllers
{
public class GameController : Controller
{
private gamezoneDBEntities db = new gamezoneDBEntities();
//
// GET: /Game/
public ViewResult Index()
{
using (var db = new gamezoneDBEntities())
{
var Info = db.tblGames.Where(UserInfo => UserInfo.UserName.Equals(User.Identity.Name)).ToList();
return View(Info);
}
}
//
// GET: /Game/Details/5
public ViewResult Details(int id)
{
tblGame tblgame = db.tblGames.Find(id);
return View(tblgame);
}
//
// GET: /Game/Create
public ActionResult Create()
{
ViewBag.ConsoleNameIDFK = new SelectList(db.tblConsoles, "ConsoleName", "ConsoleName");
return View(new tblGame { UserName = @User.Identity.Name });
}
//
// POST: /Game/Create
[HttpPost]
public ActionResult Create(tblGame tblgame)
{
if (ModelState.IsValid)
{
db.tblGames.Add(tblgame);
db.SaveChanges();
return RedirectToAction("Index");
}
ViewBag.ConsoleNameIDFK = new SelectList(db.tblConsoles, "ConsoleName", "ConsoleName", tblgame.ConsoleNameIDFK);
return View(tblgame);
}
//
// GET: /Game/Edit/5
public ActionResult Edit(int id)
{
tblGame tblgame = db.tblGames.Find(id);
ViewBag.ConsoleNameIDFK = new SelectList(db.tblConsoles, "ConsoleName", "ConsoleName", tblgame.ConsoleNameIDFK);
return View(tblgame);
}
//
// POST: /Game/Edit/5
[HttpPost]
public ActionResult Edit(tblGame tblgame)
{
if (ModelState.IsValid)
{
db.Entry(tblgame).State = EntityState.Modified;
db.SaveChanges();
return RedirectToAction("Index");
}
ViewBag.ConsoleNameIDFK = new SelectList(db.tblConsoles, "ConsoleName", "ConsoleName", tblgame.ConsoleNameIDFK);
return View(tblgame);
}
//
// GET: /Game/Delete/5
public ActionResult Delete(int id)
{
tblGame tblgame = db.tblGames.Find(id);
return View(tblgame);
}
//
// POST: /Game/Delete/5
[HttpPost, ActionName("Delete")]
public ActionResult DeleteConfirmed(int id)
{
tblGame tblgame = db.tblGames.Find(id);
db.tblGames.Remove(tblgame);
db.SaveChanges();
return RedirectToAction("Index");
}
protected override void Dispose(bool disposing)
{
db.Dispose();
base.Dispose(disposing);
}
}
}
我在索引中使用了此代码,因此每个游戏对于发布游戏的用户来说都是唯一的:
var Info = db.tblGames.Where(UserInfo => UserInfo.UserName.Equals(User.Identity.Name)).ToList();
return View(Info);
并在我的控制器的第一个创建部分添加了这个:
return View(new tblGame { UserName = @User.Identity.Name });
当您不单击创建并且每个用户都有单独的游戏帖子时,此代码确实有效,但是当我按创建新时它会崩溃
任何帮助,将不胜感激。请询问您是否需要了解更多信息