大家好!
我试图打开一个弹出式的开放用户单击GridView控件内的按钮。我使用的AJAX ModelPopUpExtender。当我运行该项目,它不显示任何错误,但它不是工作压力太大。然而,当我申请断点,它说 - :
"有没有可用于当前位置的源代码。"
请帮助我。我已经注册了工具包的情况下u要知道。
。aspx代码
<form id="form1" runat="server">
<asp:ToolkitScriptManager ID="ToolkitScriptManager1" runat="server">
</asp:ToolkitScriptManager>
<asp:UpdatePanel ID="UpdatePanel1" runat="server">
<ContentTemplate>
<asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="False"
BackColor="LightGoldenrodYellow" BorderColor="Tan" BorderWidth="1px"
CellPadding="2" DataSourceID="SqlDataSource1" ForeColor="Black"
GridLines="None" onselectedindexchanged="GridView1_SelectedIndexChanged" DataKeyNames="Name">
<Columns>
<asp:TemplateField HeaderText="Name">
<ItemTemplate>
<asp:LinkButton ID="lnkbtnName" runat="server" Text='<%# Eval("Name") %>' OnClick="lnkbtnName_Click"></asp:LinkButton>
</ItemTemplate>
</asp:TemplateField>
<asp:BoundField DataField="Company" HeaderText="Company" />
</Columns>
<FooterStyle BackColor="Tan" />
<PagerStyle BackColor="PaleGoldenrod" ForeColor="DarkSlateBlue"
HorizontalAlign="Center" />
<SelectedRowStyle BackColor="DarkSlateBlue" ForeColor="GhostWhite" />
<HeaderStyle BackColor="Tan" Font-Bold="True" />
<AlternatingRowStyle BackColor="PaleGoldenrod" />
</asp:GridView>
<br />
<br />
<asp:Button runat="server" ID="btnModalPopUp"
style="display:none"/>
<asp:ModalPopupExtender ID="ModalPopupExtender1" runat="server" TargetControlID="btnModalPopUp" PopupControlID="pnlPopUp" OkControlID="btnOk" X="60" Y="100">
</asp:ModalPopupExtender>
<asp:Panel runat="server" ID="pnlPopUp">
<asp:GridView runat="server" ID="GridView2" AutoGenerateColumns="False"
CellPadding="