如何在按钮点击时显示asp.net gridview的弹出窗口 [英] how to show pop up of asp.net gridview on button click
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问题描述
我有一个带有选择链接列和其他列的asp.net gridview,我想在按钮点击时显示gridview弹出窗口,我想将所选列id传递给页面,我该如何实现任务。 br $>
请帮帮我。
提前谢谢
I have an asp.net gridview with select link column and some other columns, i want to show pop up of gridview on button click and i want to pass the selected columns id to page, how can i achieve the task.
pls help me.
thanks in advance
推荐答案
代码为itemtemplate
Code For itemtemplate
<asp:linkbutton id="LinkButton3" runat="server" commandname="passid" commandargument="<%# Bind("Id") %>" text='Select' xmlns:asp="#unknown"></asp:linkbutton>
< br $> b $ b
protected void GridView1_RowCommand(object sender, GridViewCommandEventArgs e)
{
if (e.CommandName == "passid")
{
Session["id"] = e.CommandArgument.ToString();
//Label18.Text = Session["id"].ToString();
Response.Redirect("yourpage.aspx?id=" + Session["id"].ToString() + "");
}
}
使用aref标签在该选择按钮上
OR
use aref tag On that select button
<a href='yourpage.aspx?id=<%# Eval("Id") %>'">
<asp:Label ID="Label1" runat="server" Text="Select" ></asp:Label>
</a>
您可以通过使用事件冒泡来实现,请通过以下链接获取帮助。
HTTP:// www.aspforums.net/Threads/414279/How-to-open-Popup-Window-on-click-of-Link-Button-inside-GridView-in-ASPNet/ [ ^ ]
You can achieve by using event bubbling, go through the below link for your help.
http://www.aspforums.net/Threads/414279/How-to-open-Popup-Window-on-click-of-Link-Button-inside-GridView-in-ASPNet/[^]
hi,
你可以使用下面的代码将选定的列ID传递给页面
you can pass selected columns id to the page by using the bellow code
<asp:templatefield headertext="Select" itemstyle-horizontalalign="Center" xmlns:asp="#unknown">
<itemtemplate>
<asp:imagebutton id="imgsearch" runat="server" commandargument="<%# Eval("Id") %>" />
</itemtemplate>
<itemstyle horizontalalign="Center" />
</asp:templatefield>
aspx.cs页面中的
public static int id;
id = Convert.ToInt32(((ImageButton)sender).CommandArgument);
in aspx.cs page
public static int id;
id = Convert.ToInt32(((ImageButton)sender).CommandArgument);
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