即使另一个用户登录,current.user.identity仍然相同 [英] current.user.identity remains same even if another user logs in
本文介绍了即使另一个用户登录,current.user.identity仍然相同的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我试图根据用户ID在下拉列表中显示项目.问题是即使我使用不同的用户名登录,也选择了我登录时使用的前一个或第一个用户名.像这样.
使用
在另一个Web表单中获取用户名
字符串用户名= HttpContext.Current.User.Identity.Name;
和用户名始终保留一个人的名字..对于登录即时消息,使用:
公共 void btnlogin_Click(对象发件人,EventArgs e)
{
使用(DirectoryEntry条目= 新 DirectoryEntry())
{
// Session ["username"] = txtusername;
entry.Username = txtusername.Text;
entry.Password = txtpassword.Text;
DirectorySearcher searcher = 新 DirectorySearcher(entry);
searcher.Filter = " ;
尝试
{
searcher.FindOne();
lblrem.Text = " ;
Response.Redirect(" );
}
捕获(以前是COMException)
{
如果(例如ErrorCode == -2147023570)
{
lblrem.Text = " ;
}
}
}
解决方案
清除cookie并重试,或者在其他浏览器中尝试,您是否使用过cookie的持久状态? blockquote>
如果您没有其他浏览器,则可以尝试"InPrivate浏览"选项.
如果问题仍然存在,则需要通过调试检查代码.
问候,
Venugopal
Im trying to display items in a dropdown list according to the user id.The problem is that even though I log in with a diffrent username the previous or the very first username that i logged with is picked.I dnt know why its behaving like this.
To get the username in another web form im using
string username = HttpContext.Current.User.Identity.Name;
and username always remains of one persons name..and for the login im using:
public void btnlogin_Click(object sender, EventArgs e) { using (DirectoryEntry entry = new DirectoryEntry()) { //Session["username"] = txtusername; entry.Username = txtusername.Text; entry.Password = txtpassword.Text; DirectorySearcher searcher = new DirectorySearcher(entry); searcher.Filter = "(objectclass=user)"; try { searcher.FindOne(); lblrem.Text = "Successfull"; Response.Redirect("Default2.aspx"); } catch (COMException ex) { if (ex.ErrorCode == -2147023570) { lblrem.Text = "Login Failed.Please ensure that your username and password is correct"; } } }解决方案clear cookies and try again, or try in different browser, did u use persistant state of cookie>>>>
If you don''t have a different browser, you can try the "InPrivate Browsing" option.
And if the problem still persists you need to check your code through debugging.
Regards,
Venugopal
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