具有用户控件的ajax模式弹出窗口 [英] ajax modal popup with user control
本文介绍了具有用户控件的ajax模式弹出窗口的处理方法,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
我在asp.net中有usercontrol(ascx)
并将其放到内容页面
我想做什么
点击它后我的内容页面上有一个按钮
所以usercontrol(ascx)出现在ajax模态弹出窗口中
如何做到这一点?
解决方案
你做了
1.在页面上添加用户控件的引用。
2.在页面上添加ToolkitScriptManager。
3.你的模型弹出窗口
4.在面板中添加usercontrol
并设计如下设计
< asp:ToolkitScriptManager ID = ScriptManager1 runat = server >
< < span class =code-leadattribute> / asp:ToolkitScriptManager >
< asp:按钮 runat < span class =code-keyword> = server ID = hidden1 style = display:none / >
< asp:ModalPopupExtender ID = ModalPopupExtender1 runat = < span class =code-keyword> server
PopupControlID = Panel1 TargetControlID = hidden1 >
< / asp:ModalPopupExtender >
< asp:Panel ID = Panel1 runat = server BackColor = Aqua >
/ *这里添加你的用户控件* /
< br / >
< / asp:Panel >
嗨亲爱的请点这个网址。
mpleSite /> http://www.asp.net/ajaxlibrary/AjaxControlToolkitSampleSite/ [ ^ ] [ ]
http:// asp.net.bigresource.com/AJAX-Using-ModalPopupExtender-with-User-Control--nsmV04uO4.html [ ^ ] [ ]
i have usercontrol(ascx) in asp.net
and place it to content page
what i want to do
there is a button on my content page when click on it
so usercontrol(ascx) appear in ajax modal popup
how to do this ?
解决方案
you have do
1.Add reference of user control on page.
2.Add ToolkitScriptManager on page.
3.Your model popup
4.In panel add usercontrol
and do designing something like as follow
<asp:ToolkitScriptManager ID="ScriptManager1" runat="server"> </asp:ToolkitScriptManager> <asp:Button runat="server" ID="hidden1" style="display:none"/> <asp:ModalPopupExtender ID="ModalPopupExtender1" runat="server" PopupControlID="Panel1" TargetControlID="hidden1"> </asp:ModalPopupExtender> <asp:Panel ID="Panel1" runat="server" BackColor="Aqua"> /*here add ur user control*/ <br /> </asp:Panel>
Hi Dear please gothrow this URL.
mpleSite/">http://www.asp.net/ajaxlibrary/AjaxControlToolkitSampleSite/[^][]
http://asp.net.bigresource.com/AJAX-Using-ModalPopupExtender-with-User-Control--nsmV04uO4.html[^][]
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