2012-05-25 135 views

回答

1

使用Popup Control - Content URL显示控件中的另一个webform。在按钮的客户端单击事件上打开弹出窗口。

例如:

<dx:ASPxPopupControl ID="popup" runat="server" ContentUrl="~/ContentPageWithTextBox.aspx" 
     Top="100" ClientInstanceName="clientPopup" CloseAction="CloseButton"> 
     <ClientSideEvents Shown="OnShown" /> 
    </dx:ASPxPopupControl> 

入住这search Result
参考:
How to manipulate client-side objects within a ASPxPopupControl with the specified ContentUrl - 检查例子也。

希望对您有所帮助...

0

下面是示例代码:

<input id="openBtn" type="button" value="Open popup" 
     onclick="myPopup.SetContentUrl('http://www.google.com');myPopup.Show();" />     

<dx:ASPxPopupControl runat="server" ClientInstanceName="myPopup" Modal="True" 
        CloseAction="CloseButton" Width="500px" Height="400px"/> 

ASPxClientPopupControl members
ASPxPopupControl members

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