datagrid导出html代码

最新推荐文章于 2025-05-22 11:28:32 发布

原创最新推荐文章于 2025-05-22 11:28:32 发布 · 213 阅读

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#HTML #Excel #ASP.net #IE #UI

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本文介绍了一种将ASP.NET中的Datagrid数据导出到Excel文件的技术方案。该方法不适用于数据交换，并且Datagrid无法实现分页或排序等功能。文章提供了具体的实现步骤和代码示例。

局限性：
1、不适合数据交换，里面有html代码，比较乱，没有固定格式
2、datagrid不能有分页、排序等，否则出错

实现方法：
Response.Clear();
???Response.Buffer= false;
???Response.Charset="GB2312";
???Response.AppendHeader("Content-Disposition","attachment;filename=test.xls");
???Response.ContentEncoding=System.Text.Encoding.GetEncoding("GB2312"); ???Response.ContentType = "application/ms-excel"; ???this.EnableViewState = false;
???System.IO.StringWriter oStringWriter = new System.IO.StringWriter();
???System.Web.UI.HtmlTextWriter oHtmlTextWriter = new System.Web.UI.HtmlTextWriter(oStringWriter);
???this.DataGrid1.RenderControl(oHtmlTextWriter);
???Response.Write(oStringWriter.ToString());
???Response.End();

ASP.NET直接下?一个文件，而不是在IE中打?它

.??各位大侠，用以下?句：
context.Response.ContentType = "application/octet-stream";
context.Response.AddHeader("Content-Disposition", "attachment;FileName=" + name);
byte[] fileBuffer = new byte[fileSize];
fileStream.Read(fileBuffer, 0, (int)fileSize);
context.Response.BinaryWrite(fileBuffer);
context.Response.End();
======================================

FileStream fileStream=new FileStream("C:\\doc\\test.doc",FileMode.Open);
long fileSize = fileStream.Length;
Context.Response.ContentType="application/octet-stream";
Context.Response.AddHeader("Content-Disposition","attachment; filename=\"" + HttpUtility.UrlEncode(fileName,System.Text.Encoding.UTF8) + "\"");
Context.Response.AddHeader("Content-Length",fileSize.ToString());
byte[] fileBuffer=new byte[fileSize];
fileStream.Read(fileBuffer, 0, (int)fileSize);
fileStream.Close();
Context.Response.BinaryWrite(fileBuffer);
Context.Response.End();

============================================
int id = int.Parse(context.Request.QueryString["id"]);
            string sendUserNo = context.Request.QueryString["sendUserNo"];
            string receiveUserNo = context.Request.QueryString["receiveUserNo"];
            Message message = new Message(id, sendUserNo, receiveUserNo);

string name = context.Server.UrlEncode(message.FileName);

            string oppositeName = message.Filepath;
            string absoluteName = context.Server.MapPath(message.Filepath);
            FileStream fileStream = new FileStream(absoluteName, FileMode.Open,FileAccess.Read);
            long fileSize = fileStream.Length;

            context.Response.ContentType = "application/octet-stream";
            context.Response.AddHeader("Content-Disposition", "attachment;FileName=" + name);
            byte[] fileBuffer = new byte[fileSize];
            fileStream.Read(fileBuffer, 0, (int)fileSize);
            context.Response.BinaryWrite(fileBuffer);
            context.Response.End();