Question

alt text 您好我在ASP.net页面中使用更新面板进行部分页面回发。我有gridview和一个按钮“导出”，点击它将从下面的类调用导出方法。

如果我删除更新面板，一切都会按预期工作。但是，如果有更新面板，它将无法正常工作并抛出此错误。

请帮忙。以下是

的代码

using System;
using System.Data;
using System.Configuration;
using System.IO;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Web.UI.HtmlControls;

/// <summary>
/// 
/// </summary>
public class GridViewExportUtil
{
    /// <summary>
    /// 
    /// </summary>
    /// <param name="fileName"></param>
    /// <param name="gv"></param>
    public static void Export(string fileName, GridView gv)
    {
        HttpContext.Current.Response.Clear();
        HttpContext.Current.Response.AddHeader(
            "content-disposition", string.Format("attachment; filename={0}", fileName));
        HttpContext.Current.Response.ContentType = "application/ms-excel";

        using (StringWriter sw = new StringWriter())
        {
            using (HtmlTextWriter htw = new HtmlTextWriter(sw))
            {
                //  Create a form to contain the grid
                Table table = new Table();

                //  add the header row to the table
                if (gv.HeaderRow != null)
                {
                    GridViewExportUtil.PrepareControlForExport(gv.HeaderRow);
                    table.Rows.Add(gv.HeaderRow);
                }

                //  add each of the data rows to the table
                foreach (GridViewRow row in gv.Rows)
                {
                    GridViewExportUtil.PrepareControlForExport(row);
                    table.Rows.Add(row);
                }

                //  add the footer row to the table
                if (gv.FooterRow != null)
                {
                    GridViewExportUtil.PrepareControlForExport(gv.FooterRow);
                    table.Rows.Add(gv.FooterRow);
                }

                //  render the table into the htmlwriter
                table.RenderControl(htw);

                //  render the htmlwriter into the response
                HttpContext.Current.Response.Write(sw.ToString());
                HttpContext.Current.Response.End();
            }
        }
    }

    /// <summary>
    /// Replace any of the contained controls with literals
    /// </summary>
    /// <param name="control"></param>
    private static void PrepareControlForExport(Control control)
    {
        for (int i = 0; i < control.Controls.Count; i++)
        {
            Control current = control.Controls[i];
            if (current is LinkButton)
            {
                control.Controls.Remove(current);
                control.Controls.AddAt(i, new LiteralControl((current as LinkButton).Text));
            }
            else if (current is ImageButton)
            {
                control.Controls.Remove(current);
                control.Controls.AddAt(i, new LiteralControl((current as ImageButton).AlternateText));
            }
            else if (current is HyperLink)
            {
                control.Controls.Remove(current);
                control.Controls.AddAt(i, new LiteralControl((current as HyperLink).Text));
            }
            else if (current is DropDownList)
            {
                control.Controls.Remove(current);
                control.Controls.AddAt(i, new LiteralControl((current as DropDownList).SelectedItem.Text));
            }
            else if (current is CheckBox)
            {
                control.Controls.Remove(current);
                control.Controls.AddAt(i, new LiteralControl((current as CheckBox).Checked ? "True" : "False"));
            }

            if (current.HasControls())
            {
                GridViewExportUtil.PrepareControlForExport(current);
            }
        }
    }
}

Answer 1

在AJAX部分回发期间写入响应流时，无法正确解释响应，如果要写入响应流，则必须执行完整回发。

添加

<Triggers>
      <asp:PostBackTrigger ControlID="Your Button ID Here" />
</Triggers>

正好在

之下

</ContentTemplate>

更新小组。

更新面板和内容类型问题

1 个答案: