1

我正在构建一个应用程序,它将关闭服务器,ping 该服务器,并在服务器已关闭时通知用户。如果它没有成功关闭,它允许用户“再试一次”。我可以让应用程序第一次正确处理逻辑,但是当我实现“重试”功能时,它不会重新处理关闭操作。我认为它一定与缓存有关,但我对它的了解还不够,不知道从哪里开始寻找。有任何想法吗?

这是我的代码:

控制器

    //URL: Build/Progress  
    public ActionResult Progress()  
    {  
        return View();  
    }  

////URL: Build/MoveDevice  
public ActionResult ShutdownStatus()  
{  
    ImageInfo ImageInfo = new ImageInfo();  
    FarmInfo FarmInfo = new FarmInfo();  
    ProgressModel ProgressModel = new ProgressModel();  

    if ((Session["ShutdownStatus"] as string) == "Success")  
    {  
        ProgressModel.ShutdownStatus = 1;  
        return View(ProgressModel);  
    }  
    else  
    {  
        ImageInfo = svcImage.GetImageInfoByImageId(Session["ImageName"].ToString());  
        string Farm = ImageInfo.FarmId.ToString();  
        FarmInfo = svcFarm.GetFarmByFarmId(Farm);  
        svcImageSvc.ShutdownDevice(Session["Username"].ToString(), Session["Password"].ToString(), Session["Domain"].ToString(),  
            FarmInfo.farmName, ImageInfo.Device, ImageInfo.ImageHistory.Status, ImageInfo.PVSHost, ImageInfo.PVSAccount);  

        ProgressModel.ShutdownStatus = svcImageSvc.PingDevice(ImageInfo.Device);  
        return View(ProgressModel);  
    }  
}

视图 (Progress.aspx)

<%@ Page Title="" Language="C#" MasterPageFile="~/Views/Shared/Site.Master" Inherits="System.Web.Mvc.ViewPage<dynamic>" %>  

<asp:Content ID="Content1" ContentPlaceHolderID="TitleContent" runat="server">  
    Progress  
</asp:Content>  

<asp:Content ID="Content5" ContentPlaceHolderID="HeadContent" runat="server">  

</asp:Content>  

<asp:Content ID="Content2" ContentPlaceHolderID="MainContent" runat="server">  
    <h1>Progress</h1>  
    <hr />  

    <!-- Shutdown Status -->  
    <div class="panel" id="panel1"><img src="../../Assets/Images/load.gif" /></div>  
    <script type="text/javascript">  
        $.get('<%= Url.Action("ShutdownStatus", "Build", null) %>',  
        function (data) {  
            $('#panel1').replaceWith(data);  
        });  
    </script>  

</asp:Content>  

<asp:Content ID="Content3" ContentPlaceHolderID="HeaderContent" runat="server">  
</asp:Content>  

<asp:Content ID="Content4" ContentPlaceHolderID="FooterContent" runat="server">  
</asp:Content>

查看 (ShutdownStatus.asxc)

<%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<PvsUtil.App.Core.Models.ProgressModel>" %>  

    <% using (Html.BeginForm())  
       { %>  
        <% if (Model.ShutdownStatus == 1)  
           { %>  
           <%: Session["ShutdownStatus"] = "Success" %>  
            <p>  
                Server shutdown succesfully  
            </p>  
            <% } %>  

        <% if (Model.ShutdownStatus == 2)  
            { %>  
            <p>Server did not shutdown within 1 minute.  Please check the server and try again.</p>  
            <%: Html.ActionLink("Try Again", "Progress", "Build") %>  
            <% } %>  

        <% if (Model.ShutdownStatus == 3)  
        { %>  
        <p>An error has occurred.  Please check the server and try again.</p>  
        <%: Html.ActionLink("Try Again", "Progress", "Build") %>  
        <% } %>  
     <% } %>  
4

2 回答 2

3

我相信缓存实际上是由 jQuery 完成的。将 $.get 更改为 $.post 并查看它是否有效。如果是这样,您仍然可以使用 $.get,但您需要设置选项:

cache: false

HTH,布赖恩

于 2010-12-02T20:52:47.637 回答
0

尝试添加:

[OutputCache( Location = OutputCacheLocation.None )]

到您的控制器(或您不想被缓存的每个操作)。

您要做的另一件事是将您的 javascript 包装在 document.ready() 处理程序中。这并不是绝对必要的,因为您的脚本出现在它引用的元素之后,但它仍然是一个很好的做法——您也可以将它移动到结束 body 标记的上方,但这更像是一种优化。

<script type="text/javascript">
  $(function() { // execute on document ready   
      $.get('<%= Url.Action("ShutdownStatus", "Build", null) %>',   
      function (data) {   
          $('#panel1').replaceWith(data);   
      });
  });  
</script>
于 2010-12-02T20:48:42.083 回答