问题描述

ASP.NET文件上传问题，一上传就提示失败

using System;
using System.Data;
using System.Configuration;
using System.Collections;
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;

public partial class xueshengzuoye: System.Web.UI.Page
{
public static string fname;
protected void Page_Load(object sender EventArgs e)
{
if (!IsPostBack)
{

        if (Request[""ID""] != null)        {            InitData();        }    }}private void FileUpLoad(){    if (UploadFile.Value != null)    {        string nam = UploadFile.PostedFile.FileName;        //取得文件名(抱括路径)里最后一个"".""的索引        int i = nam.LastIndexOf(""."");        //取得文件扩展名        string newext = nam.Substring(i);        //这里我自动根据日期和文件大小不同为文件命名确保文件名不重复        DateTime now = DateTime.Now;        string newname = now.ToFileTimeUtc() + UploadFile.PostedFile.ContentLength.ToString();        //保存文件到你所要的目录这里是IIS根目录下的uploadfiles目录        //注意: 我这里用Server.MapPath()取当前文件的绝对目录.在asp.net里""""必须用""\""代替        UploadFile.PostedFile.SaveAs(Server.MapPath(""UpLoad/"" + newname + newext));        //得到这个文件的相关属性:文件名文件类型文件大小        fname = ""UpLoad/"" + newname + newext;    }}protected void InitData(){    string id = Request[""ID""].ToString();    try    {        string SQL = ""select * from TB_xueshengzuoye where ID='"" + id + ""'"";        DataTable dt = DBHelper.GetDataSet(SQL).Tables[0];        if (dt.Rows.Count > 0)        {            this.TextBox1.Text = dt.Rows[0][""ZuoYeName""].ToString();        }    }    catch    {    }}protected void Button1_Click(object sender EventArgs e){    if (UploadFile.Value != null && UploadFile.Value != """")    {        FileUpLoad();    }    string SQL = """";    bool iRet = false;    if (Request[""ID""] != null)    {        string id = Request[""ID""].ToString();        SQL = ""update TB_xueshengzuoye set ZuoYeName='"" + this.TextBox1.Text + ""'URl='"" + fname + ""' where ID='"" + id + ""'"";        iRet = DBHelper.ExecSql(SQL);    }    else    {        SQL = ""insert into TB_xueshengzuoye(ZuoYeNameURlAddUser ) values('"" + this.TextBox1.Text + ""''"" + fname + ""''"" + Session[""username""].ToString() + ""')"";        iRet = DBHelper.ExecSql(SQL);    }    if (iRet)    {        Response.Write(""<script>alert('恭喜您，操作成功！');document.location=document.location;</script>"");    }    else    {        Response.Write(""<script>alert('对不起，操作失败！');document.location=document.location;</script>"");    }}protected void Button2_Click(object sender EventArgs e){    this.TextBox1.Text = """";}

}
为什么一上传文件就提示对不起，操作失败！

解决方案

一个办法，代码跟踪一下

解决方案二：
自己设置断点跟踪调试一下，看下是文件上传失败还是数据库记录失败。

解决方案三：
建议如下:
1.文件保存与写入数据库分开调试
2.注意文件保存文件名不能相同
3.可能存在aspx执行文件路径与保存文件路径的匹派问题.以下VB代码供参考:

       Dim oleDbConnection11 As System.Data.OleDb.OleDbConnection    Dim oleDbDataAdapter11 As System.Data.OleDb.OleDbDataAdapter    Dim oleDbCommand11 As System.Data.OleDb.OleDbCommand    Dim reader_one As System.Data.OleDb.OleDbDataReader    Dim temp As String    oleDbConnection11 = New OleDbConnection    oleDbDataAdapter11 = New OleDbDataAdapter    oleDbCommand11 = New OleDbCommand    oleDbConnection11.ConnectionString = Session(""material_connectionstring"")    oleDbCommand11.Connection = oleDbConnection11    If Len(FileUpload1.PostedFile.FileName.Trim) = 0 Then        Page.ClientScript.RegisterClientScriptBlock(Me.GetType()PopupScript""alert('请选择一个上传文件!');"" True)        Return    End If    Dim mis kind file_path file_name upload_name file_add As String    Dim upload_id index_value file_length tmp As Double    mis = Request.QueryString(""mis"")    kind = Request.QueryString(""kind"")    file_add = Request.QueryString(""file_add"")    upload_id = Request.QueryString(""upload_id"")

重点关注 ** file_path = Left(Request.PhysicalPath InStrRev(Request.PhysicalPath"" -1) - 1) + ""upload"" **

    file_length = FileUpload1.PostedFile.ContentLength    tmp = InStrRev(FileUpload1.PostedFile.FileName"")    upload_name = Mid(FileUpload1.PostedFile.FileName tmp + 1 Len(FileUpload1.PostedFile.FileName) - tmp + 1)    ''返回    If Len(FileUpload1.PostedFile.FileName) = 0 Then        Page.ClientScript.RegisterClientScriptBlock(Me.GetType()PopupScript""alert('没有选择上传文件请检查!');"" True)        Return    End If    oleDbConnection11.Open()    '取最大序号    temp = ""select max(t.index_value) from upload t where t.mis = '""    temp = temp + mis + ""' and t.kind = '""    temp = temp + kind + ""' and t.upload_id = ""    temp = temp + upload_id.ToString    'TextBox1.Text = temp    'Return    oleDbCommand11.CommandText = temp    reader_one = oleDbCommand11.ExecuteReader()    Do While reader_one.Read()        If reader_one.IsDBNull(0) Then            index_value = 1        Else            index_value = (reader_one.GetValue(0) + 1)        End If    Loop    reader_one.Close()    oleDbConnection11.Close()    file_name = mis + ""_"" + kind + ""_"" + upload_id.ToString + ""_"" + index_value.ToString + ""."" + Right(upload_name.Trim (upload_name.Trim.Length - InStr(upload_name.Trim."")))    '写入数据库    oleDbConnection11.Open()    temp = ""INSERT INTO upload(mis kind upload_id index_valuefile_lengthupload_namefile_path file_name) VALUES('""    temp = temp + mis + ""'  '""    temp = temp + kind + ""'     temp = temp + upload_id.ToString + ""     temp = temp + index_value.ToString + ""     temp = temp + file_length.ToString + ""  '""    temp = temp + upload_name + ""'  '""    temp = temp + file_path + ""'  '""    temp = temp + file_name + ""')""    oleDbCommand11.CommandText = temp    oleDbCommand11.ExecuteNonQuery()    oleDbConnection11.Close()    '保存文件                           FileUpload1.PostedFile.SaveAs(file_path + file_name)    upload_status.Value = ""1""    'Label1.Text = file_name

解决方案四：
错误是什么都不贴出来，谁知道你报什么呀，还有你要注间是不是超大了！

解决方案五：
同上，跟踪代码啊，兄弟勤快些，耐心些哦