<asp:DataGrid RunAt="server" Id="myGrid" OnUpdateCommand="myGrid_Save">
<asp:TemplateColumn>
<ItemTemplate>
<asp:Label RunAt="server"
Text='<%#
DataBinder.Eval(Container.DataItem, "FileName") %>' />
</ItemTemplate>
<EditItemTemplate>
<asp:HtmlInputFile RunAt="server" Id="myFile" />
</EditItemTemplate>
</asp:TemplateColumn>
</asp:DataGrid> In your code behind file, you can use the DataGridCommandEventArgs
Item.FindControl("myFile") and cast it as a
System.Web.UI.HtmlControls.HtmlInputFile:protected void Save(DataGridCommandEventArgs e)
{
HttpPostedFile someFile =
(HtmlInputFile)e.Item.FindControl("myFile");
someFile.SaveAs(Server.MapPath("/") + "\somedir\filename.gif");
}
<asp:TemplateColumn>
<ItemTemplate>
<asp:Label RunAt="server"
Text='<%#
DataBinder.Eval(Container.DataItem, "FileName") %>' />
</ItemTemplate>
<EditItemTemplate>
<asp:HtmlInputFile RunAt="server" Id="myFile" />
</EditItemTemplate>
</asp:TemplateColumn>
</asp:DataGrid> In your code behind file, you can use the DataGridCommandEventArgs
Item.FindControl("myFile") and cast it as a
System.Web.UI.HtmlControls.HtmlInputFile:protected void Save(DataGridCommandEventArgs e)
{
HttpPostedFile someFile =
(HtmlInputFile)e.Item.FindControl("myFile");
someFile.SaveAs(Server.MapPath("/") + "\somedir\filename.gif");
}
解决方案 »
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