ajax - mvc 部分 View ajax 更新将部分 View 作为页面返回

Question

我有部分观点认为，将项目添加到数据库后需要更新。

Index.cshtml:

@using (Ajax.BeginForm("Index", "WinEntry", new AjaxOptions { HttpMethod = "POST",      UpdateTargetId = "wingrid", InsertionMode = InsertionMode.Replace}))
{
    @Html.Partial("_winVenue")
    @Html.Partial("_singleWin")
}
<div id="wingrid">
    @Html.Partial("_wingrid")
</div>

_singleWin 有提交按钮

Controller :

[HttpPost]
public ActionResult Index(Win win)
{
    win.dealerId = "1234567890";
    win.posterid = "chris";
    win.posttime = DateTime.Now;
    wem.addWin(win);
    IEnumerable<Win> w = wem.getVenueWins(win.venue,win.windate);
    return PartialView("_wingrid",w);
}

当 Controller 返回局部 View _wingrid 时，它会将其作为新页面返回，我正在寻找的行为就像是 wingrid div 内的更新面板。

如有任何帮助，我们将不胜感激。

Answer 1

由于 AJAX 表单中的 UpdateTargetId = "wingrid" 选项，您似乎已经在这样做了。只需确保从模型状态中清除在 POST Controller 操作中修改的值。否则 HTML 助手仍然可以使用旧值:

[HttpPost]
public ActionResult Index(Win win)
{
    ModelState.Remove("dealerId");
    win.dealerId = "1234567890";
    ModelState.Remove("posterid");
    win.posterid = "chris";
    ModelState.Remove("posttime");
    win.posttime = DateTime.Now;
    wem.addWin(win);
    IEnumerable<Win> w = wem.getVenueWins(win.venue,win.windate);
    return PartialView("_wingrid",w);
}

另外，如果您想让 Ajax.* 助手正常工作，请不要忘记在您的页面中包含 jquery.unobtrusive-ajax.js 脚本:

<script src="@Url.Content("~/Scripts/jquery.unobtrusive-ajax.js")" type="text/javascript"></script>