asp.net -会话-多个浏览器标签-不同的会话?

时间:2020-12-19 16:48:50

I'd like to maintain a session state per browser tab.

我想维护每个浏览器标签的会话状态。

Is this easy (or even possible) to do in ASP.NET?

这在ASP.NET中很容易(甚至可能)实现吗?

Example: A user hits Ctrl-T in firefox 5 times and visits the site in each tab. I'd like each tab to have its own session state on the server

示例:用户在firefox中按下Ctrl-T 5次,并在每个选项卡中访问站点。我希望每个选项卡在服务器上有自己的会话状态。

3 个解决方案

#1


21  

To facilitate multi-tab session states for one user without cluttering up the URL, do the following.

要为一个用户提供多选项卡会话状态,而又不影响URL,请执行以下操作。

In your form load function, include:

在表单加载函数中,包括:

If Not IsPostback Then
  'Generate a new PageiD'
  ViewState("_PageID") = (New Random()).Next().ToString()
End If

When you save something to your Session State, include the PageID:

当你保存一些东西到你的会话状态时,包括PageID:

Session(ViewState("_PageID").ToString() & "CheckBoxes") = D

Notes:

注:

  • As with session ID's in general, you cannot trust that malicious viewers will not change the SessionID / PageID. This is only a valid solution for an environment where all users can be trusted. Fortunately, ViewState does offer more protection than using a hidden input field.
  • 与会话ID一般,您不能相信恶意的查看器不会改变SessionID / PageID。对于所有用户都可以信任的环境,这只是一个有效的解决方案。幸运的是,ViewState比使用隐藏的输入字段提供更多的保护。
  • You will not have access to the PageID until the ViewState is restored upon PostBack. Therefore, you will not have access to the PageID in the page_init() handler.
  • 在回发恢复ViewState之前,您不能访问PageID。因此,您将无法访问page_init()处理程序中的PageID。

#2


15  

<configuration>
  <system.web>
    <sessionState cookieless="true"
      regenerateExpiredSessionId="true" />
  </system.web>
</configuration>

http://msdn.microsoft.com/en-us/library/ms178581.aspx

http://msdn.microsoft.com/en-us/library/ms178581.aspx

in this case each tab will get unique ID and it will looks like it is another visitor.

在这种情况下,每个选项卡都有唯一的ID,看起来就像另一个访问者。

#3


1  

Using Brian Webster's answer I found a problem with XMLHttpRequests. It turned out, XMLHttpRequests did not set the IsPostback flag to true and therefore the request looked like a new request and one would end up having a new session state for that request. To solve that problem I also checked the value of the ViewState("_PageID")

使用Brian Webster的答案,我发现了一个xmlhttprequest问题。事实证明,xmlhttprequest并没有将IsPostback标记设置为true,因此这个请求看起来就像一个新的请求,一个请求最后会有一个新的会话状态。为了解决这个问题,我还检查了ViewState的值(“_PageID”)

so that my code looks like this in C#:

我的代码在c#中是这样的

protected dynamic sessionVar; //a copy of the session variable

protected void Page_Load(object sender, EventArgs e)
{
    if (!IsPostBack && ViewState["_PageID"] == null)
    {
        ViewState["_PageID"] = (new Random()).Next().ToString();
        Session[ViewState["_PageID"] + "sessionVar"] = initSessionVar(); //this function should initialize the session variable
    }
    sessionVar = Session[ViewState["_PageID"] + "sessionVar"];
    //...
}

#1


21  

To facilitate multi-tab session states for one user without cluttering up the URL, do the following.

要为一个用户提供多选项卡会话状态,而又不影响URL,请执行以下操作。

In your form load function, include:

在表单加载函数中,包括:

If Not IsPostback Then
  'Generate a new PageiD'
  ViewState("_PageID") = (New Random()).Next().ToString()
End If

When you save something to your Session State, include the PageID:

当你保存一些东西到你的会话状态时,包括PageID:

Session(ViewState("_PageID").ToString() & "CheckBoxes") = D

Notes:

注:

  • As with session ID's in general, you cannot trust that malicious viewers will not change the SessionID / PageID. This is only a valid solution for an environment where all users can be trusted. Fortunately, ViewState does offer more protection than using a hidden input field.
  • 与会话ID一般,您不能相信恶意的查看器不会改变SessionID / PageID。对于所有用户都可以信任的环境,这只是一个有效的解决方案。幸运的是,ViewState比使用隐藏的输入字段提供更多的保护。
  • You will not have access to the PageID until the ViewState is restored upon PostBack. Therefore, you will not have access to the PageID in the page_init() handler.
  • 在回发恢复ViewState之前,您不能访问PageID。因此,您将无法访问page_init()处理程序中的PageID。

#2


15  

<configuration>
  <system.web>
    <sessionState cookieless="true"
      regenerateExpiredSessionId="true" />
  </system.web>
</configuration>

http://msdn.microsoft.com/en-us/library/ms178581.aspx

http://msdn.microsoft.com/en-us/library/ms178581.aspx

in this case each tab will get unique ID and it will looks like it is another visitor.

在这种情况下,每个选项卡都有唯一的ID,看起来就像另一个访问者。

#3


1  

Using Brian Webster's answer I found a problem with XMLHttpRequests. It turned out, XMLHttpRequests did not set the IsPostback flag to true and therefore the request looked like a new request and one would end up having a new session state for that request. To solve that problem I also checked the value of the ViewState("_PageID")

使用Brian Webster的答案,我发现了一个xmlhttprequest问题。事实证明,xmlhttprequest并没有将IsPostback标记设置为true,因此这个请求看起来就像一个新的请求,一个请求最后会有一个新的会话状态。为了解决这个问题,我还检查了ViewState的值(“_PageID”)

so that my code looks like this in C#:

我的代码在c#中是这样的

protected dynamic sessionVar; //a copy of the session variable

protected void Page_Load(object sender, EventArgs e)
{
    if (!IsPostBack && ViewState["_PageID"] == null)
    {
        ViewState["_PageID"] = (new Random()).Next().ToString();
        Session[ViewState["_PageID"] + "sessionVar"] = initSessionVar(); //this function should initialize the session variable
    }
    sessionVar = Session[ViewState["_PageID"] + "sessionVar"];
    //...
}