用ftp上传文件,下面是MSDN上的代码;public static void Main ()
{
// Get the object used to communicate with the server.
FtpWebRequest request = (FtpWebRequest)WebRequest.Create("ftp://www.contoso.com/test.htm");
request.Method = WebRequestMethods.Ftp.UploadFile; // This example assumes the FTP site uses anonymous logon.
request.Credentials = new NetworkCredential ("anonymous","[email protected]");
// Copy the contents of the file to the request stream.
StreamReader sourceStream = new StreamReader("testfile.txt");
byte [] fileContents = Encoding.UTF8.GetBytes(sourceStream.ReadToEnd());
sourceStream.Close();
request.ContentLength = fileContents.Length; Stream requestStream = request.GetRequestStream();
requestStream.Write(fileContents, 0, fileContents.Length);
requestStream.Close(); FtpWebResponse response = (FtpWebResponse)request.GetResponse();
Console.WriteLine("Upload File Complete, status {0}", response.StatusDescription);
response.Close();
}
可是在FtpWebResponse response = (FtpWebResponse)request.GetResponse();
时会出现“基础连接已经关闭: 接收时发生错误”;winform程序,这是什么原因?
{
// Get the object used to communicate with the server.
FtpWebRequest request = (FtpWebRequest)WebRequest.Create("ftp://www.contoso.com/test.htm");
request.Method = WebRequestMethods.Ftp.UploadFile; // This example assumes the FTP site uses anonymous logon.
request.Credentials = new NetworkCredential ("anonymous","[email protected]");
// Copy the contents of the file to the request stream.
StreamReader sourceStream = new StreamReader("testfile.txt");
byte [] fileContents = Encoding.UTF8.GetBytes(sourceStream.ReadToEnd());
sourceStream.Close();
request.ContentLength = fileContents.Length; Stream requestStream = request.GetRequestStream();
requestStream.Write(fileContents, 0, fileContents.Length);
requestStream.Close(); FtpWebResponse response = (FtpWebResponse)request.GetResponse();
Console.WriteLine("Upload File Complete, status {0}", response.StatusDescription);
response.Close();
}
可是在FtpWebResponse response = (FtpWebResponse)request.GetResponse();
时会出现“基础连接已经关闭: 接收时发生错误”;winform程序,这是什么原因?
这句话有问题吧,一般是这样的
string url="ftp://192.168.1.1/a.txt";
FtpWebRequest request = (FtpWebRequest)WebRequest.Create(url);
/// 利用FTP上传
/// by minjiang 07-09-05
/// </summary>
/// <param name="filename">上传的文件名</param>
private void mFtpUpload(string filename )
{
string strfileoldname;//图片保存后名称
//为文件命名,然后保存
string fileExtension;
fileExtension = System.IO.Path.GetExtension(filename);
Random ra = new Random(); strfileoldname = ra.Next(100000, 999999).ToString() + fileExtension;
FileInfo fileInf = new FileInfo(filename);
//string uri = "ftp://" + ftpServerIP + "/" + fileInf.Name;
string uri = "ftp://" + ftpServerIP + "/" + strfileoldname ;
FtpWebRequest reqFTP; // 根据uri创建FtpWebRequest对象
reqFTP = (FtpWebRequest)FtpWebRequest.Create(new Uri(uri));
//重命名文件的新名称 by minjiang 07-09-05
reqFTP.RenameTo = strfileoldname;
// ftp用户名和密码
reqFTP.Credentials = new NetworkCredential(ftpUserID, ftpPassword); // 默认为true,连接不会被关闭
// 在一个命令之后被执行
reqFTP.KeepAlive = false; // 指定执行什么命令
reqFTP.Method = WebRequestMethods.Ftp.UploadFile; // 指定数据传输类型
reqFTP.UseBinary = true; // 上传文件时通知服务器文件的大小
reqFTP.ContentLength = fileInf.Length; // 缓冲大小设置为2kb
int buffLength = 2048; byte[] buff = new byte[buffLength];
int contentLen;
//设置上传进度变量
double dProgess = 1.00;
// 打开一个文件流 (System.IO.FileStream) 去读上传的文件
FileStream fs = fileInf.OpenRead();
FtpWebResponse resFTP = null ;
try
{
// 把上传的文件写入流
Stream strm = reqFTP.GetRequestStream();
// 每次读文件流的2kb
contentLen = fs.Read(buff, 0, buffLength);
// 流内容没有结束
while (contentLen != 0)
{
// 把内容从file stream 写入 upload stream
strm.Write(buff, 0, contentLen); contentLen = fs.Read(buff, 0, buffLength);
}
// 关闭两个流
strm.Close();
fs.Close();
Response.Write("上传成功,文件路径:"+uri );
}
catch (Exception ex)
{
string err = ex .Message .ToString ();
if (resFTP != null)
{
string sResFTP = resFTP.StatusCode.ToString();
}
Response.Write(err);
}
}
//执行上传功能
protected void txtUpload_Click(object sender, EventArgs e)
{
HttpFileCollection files = HttpContext.Current.Request.Files;
//要上传的文件名
string fileName = this.File1.PostedFile.FileName;
this.mFtpUpload(fileName ); }