从服务器端代码打开弹出窗口

问题描述

大家好，..
在检查了一些条件后，我想从服务器端打开一个模式弹出窗口.
我尝试了许多代码，但是它们没有用.
这是我现在正在使用的..

Hello guys..,
I want to open a modal popup from server side after checking some condtions.
I tried many codes but they didn''t work.
Here''s what I am using now ..

string strScript;
        Type jsType = ObjPage.GetType();
        ClientScriptManager CS = ObjPage.ClientScript;
        strScript = "javascript:window.showModalDialog(''ChangePassword.aspx'');";
        System.Web.UI.ScriptManager.RegisterClientScriptBlock(this, jsType, "LoginButton", strScript, true);

任何帮助将不胜感激....

Any help will be appreciated....

推荐答案

<asp:ModalPopupExtender ID="mpeDemo" BehaviorID="mpebDemo" runat="server" TargetControlID="hylPopup"
                    PopupControlID="Panel1" BackgroundCssClass="popupwrapper_transparentbg" DropShadow="false"
                    RepositionMode="RepositionOnWindowScroll" PopupDragHandleControlID="pnlConnectDragHandle"
                    OkControlID="lbtnCancel" />

请从服务器上执行以下操作:

Please do the following from Server:

Button1_Click(object sender, EventArgs e)
{
   //some login write here...
   mpeDemo.Show(); //then open the AJAX Modal Window
}

如有任何疑问，请让我知道.

如果对您有帮助，请提供投票.

谢谢，
Imdadhusen

Please do let me know, if you have any doubt.

Please provide Vote if this would be helpful to you.

Thanks,
Imdadhusen

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