Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
463 views
in Technique[技术] by (71.8m points)

c# - CookieContainer bug?

I'm confused how CookieContainer handles domain, so I create this test. This test shows cookieContainer doesn't return any cookie for "example.com" but according to RFC it should return at least 2 cookies.

Isn't it a bug?

How make it to work?

Here is a discussion about this bug:

http://social.msdn.microsoft.com/Forums/en-US/ncl/thread/c4edc965-2dc2-4724-8f08-68815cf1dce6

<%@ Page Language="C#" %>

<%@ Import Namespace="System.Net" %>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<script runat="server">
    CookieContainer getContainer()
    {
        CookieContainer result = new CookieContainer();

        Uri uri = new Uri("http://sub.example.com");
        string cookieH = @"Test1=val; domain=sub.example.com; path=/";
        result.SetCookies(uri, cookieH);

        cookieH = @"Test2=val; domain=.example.com; path=/";
        result.SetCookies(uri, cookieH);

        cookieH = @"Test3=val; domain=example.com; path=/";
        result.SetCookies(uri, cookieH);

        return result;
    }

    void Test()
    {
        CookieContainer cookie = getContainer();
        lblResult.Text += "<br>Total cookies count: " + cookie.Count + " &nbsp;&nbsp; expected: 3";

        Uri uri = new Uri("http://sub.example.com");
        CookieCollection coll = cookie.GetCookies(uri);
        lblResult.Text += "<br>For " + uri + " Cookie count: " + coll.Count + " &nbsp;&nbsp; expected: 2";

        uri = new Uri("http://other.example.com");
        coll = cookie.GetCookies(uri);
        lblResult.Text += "<br>For " + uri + " Cookie count: " + coll.Count + " &nbsp;&nbsp; expected: 2";

        uri = new Uri("http://example.com");
        coll = cookie.GetCookies(uri);
        lblResult.Text += "<br>For " + uri + " Cookie count: " + coll.Count + " &nbsp;&nbsp; expected: 2";

    }

    protected void Page_Load(object sender, EventArgs e)
    {
        Test();
    }
</script>

<html xmlns="http://www.w3.org/1999/xhtml">
<head runat="server">
    <title>CookieContainer Test Page</title>
</head>
<body>
    <form id="frmTest" runat="server">
    <asp:Label ID="lblResult" EnableViewState="false" runat="server"></asp:Label>
    </form>
</body>
</html>
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I just found the fix for this bug and discussed here: http://dot-net-expertise.blogspot.com/2009/10/cookiecontainer-domain-handling-bug-fix.html

Here is the solution:

  1. Don't use .Add(Cookie), Use only .Add(Uri, Cookie) method.
  2. Call BugFix_CookieDomain each time you add a cookie to the container or before you use .GetCookie or before system use the container.

    private void BugFix_CookieDomain(CookieContainer cookieContainer)
    {
        System.Type _ContainerType = typeof(CookieContainer);
        Hashtable table = (Hashtable)_ContainerType.InvokeMember("m_domainTable",
                                   System.Reflection.BindingFlags.NonPublic |
                                   System.Reflection.BindingFlags.GetField |
                                   System.Reflection.BindingFlags.Instance,
                                   null,
                                   cookieContainer,
                                   new object[] { });
        ArrayList keys = new ArrayList(table.Keys);
        foreach (string keyObj in keys)
        {
            string key = (keyObj as string);
            if (key[0] == '.')
            {
                string newKey = key.Remove(0, 1);
                table[newKey] = table[keyObj];
            }
        }
    }
    

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...