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我有一个AJAX模式popupextender,我想使用Java脚本的一些处理后面的代码年底显现。我收到消息 错误:无法获取未定义或空引用的属性'show'。ModalPopupExtender显示方法不从后台代码工作的JavaScript
<asp:Panel ID="panel1" runat="server" Visible="true" BorderColor="Black" Style="display: none">
<asp:UpdatePanel ID="uppanel1" runat="server">
<ContentTemplate>
<asp:Button ID="btn1" Text="Popup" Visible="true" runat="server" Style="display:none"/>
<ajaxToolKit:ModalPopupExtender ID="mpe1" runat="server" TargetControlID="btn1" PopupControlID="panel1" RepositionMode="None" PopupDragHandleControlID="drag1" BehaviorID="behave1"/>
<div id="div1" runat="server">
<asp:TextBox ID="txt1" runat="server" Text="Text" ></asp:TextBox>
</div>
</ContentTemplate>
</asp:UpdatePanel>
</asp:Panel>
function ShowPopUp(mpid) {
var id1 = $find(mpid);
id1.show();
}
欢迎SO!你能否为你的答案提供一些解释? – 2013-04-29 15:22:40