如何基于gridview下拉菜单中的selectedindex显示标签中的数据已更改 [英] how to display data in label based on gridview dropdown onselectedindexchanged

问题描述

我正在使用gridview
我在dropdownlist中有第一列，而column2是label.

当我选择下拉菜单时，我想根据我的gridview下拉菜单选择的索引在标签中显示我的值..

I am using gridview
I have column one in dropdownlist and column2 is label.

when i select my dropdown, i want to display my values in the lable based on my gridview dropdown selected index changed..

what event need to use for this..?

推荐答案

创建一个事件处理程序，如下所示:

受保护的void DropDownList1_SelectedIndexChanged(对象发送者，EventArgs e)
{
//您的代码
}

然后将事件处理程序附加到下拉列表，如下所示:

< asp:GridView ID ="GridView1" runat ="server">
<专栏>
< asp:TemplateField>
< ItemTemplate>
< asp:DropDownList ID ="DropDownList1" runat =服务器" AutoPostBack ="true"
onselectedindexchanged ="DropDownList1_SelectedIndexChanged">
</asp:DropDownList>
</ItemTemplate>
</asp:TemplateField>
</列>
</asp:GridView>

create a event handler like below :

protected void DropDownList1_SelectedIndexChanged(object sender, EventArgs e)
{
// your code
}

then attach the event handler to the dropdownlist like shown below:

<asp:GridView ID="GridView1" runat="server">
<Columns>
<asp:TemplateField>
<ItemTemplate>
<asp:DropDownList ID="DropDownList1" runat="server" AutoPostBack="true"
onselectedindexchanged="DropDownList1_SelectedIndexChanged">
</asp:DropDownList>
</ItemTemplate>
</asp:TemplateField>
</Columns>
</asp:GridView>

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