Question

如何在网格视图中的项目模板中添加文件上传控件。不能从gridview.how的上传控件获取文件名，以便在gridview中为每一行添加文件上传控件

<asp:GridView runat="server" ID="gridviewAddPo" AutoGenerateEditButton="false" OnRowEditing="gridviewAddPo_RowEditing" OnRowUpdating="gridviewAddPo_RowUpdating" AutoGenerateColumns="false">
  <Columns>  
        <asp:TemplateField>  
            <ItemTemplate>  
                <asp:Button ID="btn_Edit" runat="server" Text="Edit" CommandName="Edit" />  
            </ItemTemplate>  
            <EditItemTemplate>  
                <asp:Button ID="btn_Update" runat="server" Text="Update" CommandName="Update"/>  
                <asp:Button ID="btn_Cancel" runat="server" Text="Cancel" CommandName="Cancel"/>  
            </EditItemTemplate>  
        </asp:TemplateField>  


      <asp:TemplateField>
          <ItemTemplate>
            <asp:FileUpload ID="FileUpload1" runat="server" />
            <asp:Button ID="Button1" runat="server" Text="Upload" OnClick="Upload" />
          </ItemTemplate>
      </asp:TemplateField>
      <asp:TemplateField HeaderText="ID">  
                        <ItemTemplate>  
                            <asp:Label ID="lbl_ID" runat="server" Text='<%#Eval("id") %>'></asp:Label>  
                        </ItemTemplate>  
      </asp:TemplateField>  
    <Columns/>
</asp:Gridview>

背后的代码是：

protected void Upload(object sender, EventArgs e)
{
    string filename = Path.GetFileName(FileUpload1.PostedFile.FileName);//getting errrorhow ot 
    string contentType = FileUpload1.PostedFile.ContentType;
    using (Stream fs = FileUpload1.PostedFile.InputStream)
    {
        using (System.IO.BinaryReader br = new BinaryReader(fs))
        {
            byte[] bytes = br.ReadBytes((Int32)fs.Length);
            string constr = ConfigurationManager.ConnectionStrings["kernelCS"].ConnectionString;
            using (SqlConnection con = new SqlConnection(constr))
.....
}

Answer 1

将错误行替换为低于4行

Button btn = (Button)sender;
GridViewRow gvr = (GridViewRow)btn.NamingContainer;
FileUpload fu = gvr.cells[1].FindControl("FileUpload1") as FileUpload;
string filename = Path.GetFileName(fu.PostedFile.FileName);

如何在asp.net

1 个答案: