获取fileupload/html输入控件的路径 [英] getting the path of fileupload/html input control

问题描述

我有以下编码，当我手动设置文件路径时，效果很好.即

Hi ,

I have the following coding , which works well when i set path of the file manually.. i.e,

FileInfo fileInf = new FileInfo("C:\\new\\file1.zip");
    string uri = "ftp://" +
 ftpServerIP + "/" + fileInf.Name;
    FtpWebRequest reqFTP;
 
    // Create FtpWebRequest object from the Uri provided
    reqFTP =
 (FtpWebRequest)FtpWebRequest.Create(new Uri("ftp://" + ftpServerIP +
 "/" + fileInf.Name));
 
    // Provide the WebPermission Credintials
    reqFTP.Credentials = new NetworkCredential(ftpUserID, ftpPassword);
 
    // By default KeepAlive is true, where the control connection is not closed
    // after a command is executed.
    reqFTP.KeepAlive = false;
 
    // Specify the command to be executed.
    reqFTP.Method = WebRequestMethods.Ftp.UploadFile;
 
    // Specify the data transfer type.
    reqFTP.UseBinary = true;
 
    // Notify the server about the size of the uploaded file
    reqFTP.ContentLength = fileInf.Length;
 
    // The buffer size is set to 2kb
    int buffLength = 2048;
    byte[] buff = new byte[buffLength];
    int contentLen;
 
    // Opens a file stream (System.IO.FileStream) to read the file to be uploaded
    FileStream fs = fileInf.OpenRead();
 
    try
    {
        // Stream to which the file to be upload is written
        Stream strm = reqFTP.GetRequestStream();
 
        // Read from the file stream 2kb at a time
        contentLen = fs.Read(buff, 0, buffLength);
 
        // Till Stream content ends
        while (contentLen != 0)
        {
            // Write Content from the file stream to the FTP Upload Stream
            strm.Write(buff, 0, contentLen);
            contentLen = fs.Read(buff, 0, buffLength);
        }
 
        // Close the file stream and the Request Stream
        strm.Close();
        fs.Close();
    }
    catch (Exception ex)
    {
        MessageBox.Show(ex.Message, "Upload Error");
    }

在此编码中，我将路径固定为，

FileInfo fileInf = new FileInfo("C:\\ new \\ file1.zip");

但我希望用户使用fileupload或html输入控件浏览并选择文件.问题是，我无法在fileupload中获取完整路径.它仅返回文件名，上面的代码返回错误找不到文件".因此，请帮助我如何使用文件上传控制或任何其他方式来实现这一目标.


谢谢，

Emm

in this coding , i set the path as fixed as ,

FileInfo fileInf = new FileInfo("C:\\new\\file1.zip");

but i want user to browse and select the file using fileupload or html input control. the problem is , i cant get the full path in fileupload. it just returns the filename alone and the above code returns the error "couldnot find the file". so pls help me how to achieve this by using fileupload control or anyother mreans.


Thanks,

Emm

推荐答案

获取完整路径名

To get the full path name

string path = FileUpload1.PostedFile.FileName 

这可以帮助您从FileUpload控件中获取完整的文件路径 [

This may help you getting the full file path from a FileUpload control[^]

Hii



字符串MyFilePath = FileUpload.FileName.ToString();




谢谢

Hii



string MyFilePath=FileUpload.FileName.ToString();




Thankx

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