一旦我们收到它们，有没有办法处理分块请求中的块？

Question

我希望我能正确解释自己。 我有一个带有 post 方法 get_file() 的 C# Web 服务，从我的客户端，我向服务器中的该方法发出 POST 请求，传输编码分块，然后分块发送文件。

该方法在请求完成后开始（输入 0\r\n\r\n），有没有办法在我将每个块发送到服务器后处理它，而不是等待结束？

也许发送多个 POST 请求的替代方案会更好？ （但是在大约 100 个请求之后我得到一个错误）。

再次，我希望它得到正确解释。 提前致谢

Answer 1

据我了解，您希望 C# 客户端将大文件分块上传到托管您的 C# 服务的服务器。

在这种情况下，我使用以下解决方案已经有一段时间了，并且已经证明它非常健壮：

服务器端功能：

    [WebMethod]
    public void UploadFile(string FileName, byte[] buffer, long Offset, out bool UploadOK, out string msg)
    {
        Log(string.Format("Upload File {0}. Offset {1}, Bytes {2}...", FileName, Offset, buffer.Length));
        UploadOK = false;
        try
        {
            // setting the file location to be saved in the server. 
            // reading from the web.config file 
            string FilePath = Path.Combine( ConfigurationManager.AppSettings["upload_path"], FileName);

            if (Offset == 0) // new file, create an empty file
                File.Create(FilePath).Close();
            // open a file stream and write the buffer. 
            // Don't open with FileMode.Append because the transfer may wish to 
            // start a different point
            using (FileStream fs = new FileStream(FilePath, FileMode.Open,
                FileAccess.ReadWrite, FileShare.Read))
            {
                fs.Seek(Offset, SeekOrigin.Begin);
                fs.Write(buffer, 0, buffer.Length);
            }
            UploadOK = true;
            msg = "uploaded to " + FilePath;

            Log(string.Format("Sucessfully Uploaded to File {0}: {1}", FileName, msg));
        }
        catch (Exception ex)
        {
            //sending error:
            msg = "failed to upload: " + ex.Message;
            UploadOK = false;
            Log(string.Format("Failed Upload File {0}: {1}", EmlFileName, ex.Message));
        }
    }

客户端上传功能：

static void SendFile(YourWebService webservice, string filename)
    {
        Console.WriteLine("uploading file: " + filename);

        int Offset = 0; // starting offset.

        //define the chunk size
        int ChunkSize = 65536; // 64 * 1024 kb

        //define the buffer array according to the chunksize.
        byte[] Buffer = new byte[ChunkSize];
        //opening the file for read.
        FileStream fs = new FileStream(filename, FileMode.Open, FileAccess.Read);
        try
        {
            long FileSize = new FileInfo(filename).Length; // File size of file being uploaded.
            // reading the file.
            fs.Position = Offset;
            int BytesRead = 0;
            string msg = "";
            while (Offset != FileSize) // continue uploading the file chunks until offset = file size.
            {
                BytesRead = fs.Read(Buffer, 0, ChunkSize); // read the next chunk 
                // (if it exists) into the buffer. 
                // the while loop will terminate if there is nothing left to read
                // check if this is the last chunk and resize the buffer as needed 
                // to avoid sending a mostly empty buffer 
                // (could be 10Mb of 000000000000s in a large chunk)
                if (BytesRead != Buffer.Length)
                {
                    ChunkSize = BytesRead;
                    byte[] TrimmedBuffer = new byte[BytesRead];
                    Array.Copy(Buffer, TrimmedBuffer, BytesRead);
                    Buffer = TrimmedBuffer; // the trimmed buffer should become the new 'buffer'
                }
                // send this chunk to the server. it is sent as a byte[] parameter, 
                // but the client and server have been configured to encode byte[] using MTOM. 
                bool ChunkAppened = webservice.UploadFile(Path.GetFileName(filename), Buffer, Offset, out msg);
                if (!ChunkAppened)
                {
                    Console.WriteLine("failed to upload. server return error: " + msg);
                    break;
                }
                // Offset is only updated AFTER a successful send of the bytes. 
                Offset += BytesRead; // save the offset position for resume
            }
            Console.WriteLine("successfully uploaded file: " + filename);
        }
        catch (Exception ex)
        {
            Console.WriteLine("failed to upload file: " + ex.Message);
        }
        finally
        {
            fs.Close();
        }
    }