[英]C# overwrite file with empty file when trying to upload (ftp)
我正在尝试通过ftp上传文本文件,当我运行程序时,它将清空我的文件,然后上传它。 我不知道为什么...可能会覆盖它或其他内容...这是我的代码:
这是在主班
/* Create Object Instance */
ftp ftpClient = new ftp(@"ftp://sportcaffe.me", "sport***", "*****");
/* Upload a File */
ftpClient.upload("public_html/test.txt", @"C:\Users\Lazar\Desktop\test.txt");
这是ftp类功能代码:
/* Upload File */
public void upload(string remoteFile, string localFile)
{
try
{
/* Create an FTP Request */
ftpRequest = (FtpWebRequest)FtpWebRequest.Create(host + "/" + remoteFile);
/* Log in to the FTP Server with the User Name and Password Provided */
ftpRequest.Credentials = new NetworkCredential(user, pass);
/* When in doubt, use these options */
ftpRequest.UseBinary = true;
ftpRequest.UsePassive = true;
ftpRequest.KeepAlive = true;
/* Specify the Type of FTP Request */
ftpRequest.Method = WebRequestMethods.Ftp.UploadFile;
/* Establish Return Communication with the FTP Server */
ftpStream = ftpRequest.GetRequestStream();
/* Open a File Stream to Read the File for Upload */
FileStream localFileStream = new FileStream(localFile, FileMode.Create);
/* Buffer for the Downloaded Data */
byte[] byteBuffer = new byte[bufferSize];
int bytesSent = localFileStream.Read(byteBuffer, 0, bufferSize);
/* Upload the File by Sending the Buffered Data Until the Transfer is Complete */
try
{
while (bytesSent != 0)
{
ftpStream.Write(byteBuffer, 0, bytesSent);
bytesSent = localFileStream.Read(byteBuffer, 0, bufferSize);
}
}
catch (Exception ex) { MessageBox.Show(ex.ToString()); }
/* Resource Cleanup */
localFileStream.Close();
ftpStream.Close();
ftpRequest = null;
}
catch (Exception ex) { MessageBox.Show(ex.ToString()); }
return;
}
没有错误和警告...
首先,我在文件中写入一些内容并保存,然后在运行程序时,我的文件为空,并且该empy文件已上传...
我使用此代码http://www.codeproject.com/Tips/443588/Simple-Csharp-FTP-Class
FileStream localFileStream = new FileStream(localFile, FileMode.Create);
我希望我不需要太清楚地拼写出来,但是上面的行被告知要创建一个文件。 引用文档:“指定操作系统应该创建一个新文件。如果该文件已经存在,它将被覆盖。”
http://msdn.microsoft.com/zh-cn/library/system.io.filemode(v=vs.110).aspx列出了您可以在此处使用的选项列表。 FileMode.Open
应该可以满足您的需要(它将打开文件,如果文件不存在,则抛出异常)。
解决了
FileStream localFileStream =新的FileStream(localFile,FileMode.Create); 用FileMode.Open替换FileMode.Create
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