Asp.net Ajax Modelpop不工作
我在gridview中使用了ajax modelpop:Asp.net Ajax Modelpop不工作,asp.net,ajax,Asp.net,Ajax,我在gridview中使用了ajax modelpop: <asp:GridView ID="grdInbox" runat="server" CellPadding="0" AutoGenerateColumns="False" Width="100%"> <Columns> <asp:TemplateField> <ItemTemplate> <
<asp:GridView ID="grdInbox" runat="server" CellPadding="0"
AutoGenerateColumns="False" Width="100%">
<Columns>
<asp:TemplateField>
<ItemTemplate>
<asp:Button CssClass="blueButton" ID="btnReply" Text="Reply" style="background-color:#FFA600;" runat="server"/>
<cc1:ModalPopupExtender
ID="modlpopupReply" runat="server" TargetControlID="btnReply" PopupControlID="pnlReply" CancelControlID="btnCancelReply" DropShadow="true" Enabled="True" X="100" Y="100">
</cc1:ModalPopupExtender>
</ItemTemplate>
</asp:TemplateField>
</Columns>
</asp:GridView>
<asp:Panel ClientIDMode="Static" ID="pnlReply" runat="server">
<div>
<div style="font-size:20px;">Reply Message</div>
<asp:Button ID="btnSend" runat="server" Text="Send" ClientIDMode="Static" CssClass="blueButton" />
<asp:Button ID="btnCancelReply" runat="server" ClientIDMode="Static" Text="Cancel" CssClass="blueButton" style="background-color:#FFA600;"/>
</div>
</asp:Panel>
回复信息
问题是:
- 它不会出现在屏幕上
- 如何在弹出窗口中传递gridview其他列的值