我在GridView中使用模态弹出作为LinkButton。
我需要的是使用点击的LinkButton 的值出现在弹出式面板上的标签上
答案 0 :(得分:2)
<asp:GridView ID="GridView1" runat="server">
<Columns><asp:TemplateField>
<ItemTemplate>
<asp:LinkButton ID="LinkButton1" runat="server" OnClick="LinkButton1_Click" Text='<%#Eval("empname") %>'></asp:LinkButton>
</ItemTemplate>
</asp:TemplateField></Columns>
</asp:GridView>
<asp:LinkButton ID="lnkDummy" runat="server"></asp:LinkButton>
<asp:ModalPopupExtender ID="LinkButton1_ModalPopupExtender" runat="server" DynamicServicePath="" Enabled="True" TargetControlID="lnkDummy" PopupControlID="Panel1"> </asp:ModalPopupExtender>
<asp:Panel ID="Panel1" runat="server" Height="164px" Width="284px" BackColor="#CCCCFF" ><br /><br />
<center><asp:Label ID="Label2" runat="server" Text="Label"></asp:Label><br /><br />
<asp:TextBox ID="TextBox1" runat="server"></asp:TextBox><br /><br />
<asp:Button ID="Button1" runat="server" Text="Do"/></center>
</asp:Panel>
码
protected void LinkButton1_Click(object sender, EventArgs e)
{
LinkButton Lnk = (LinkButton)sender;
Label2.Text = Lnk.Text;
LinkButton1_ModalPopupExtender.Show();
}
答案 1 :(得分:0)
LinkButton btndetails = (LinkButton)e.CommandSource;
您现在可以访问您希望显示的btndetails文本。