c# - 带语音的 Lex 聊天机器人 C# 客户端

Question

我让聊天机器人客户端运行文本，但现在想将其更改为语音，但我不确定如何从麦克风获取流以进行发布。对于录制音频，我正在使用 NAudio，但是在发送内存流时，我收到一条错误消息

System.IO.IOException：在写入所有字节之前无法关闭流。

这是我的代码：

private void recordAudio()
        {
            if (memoryStream == null)
                memoryStream = new MemoryStream();
            sourceStream = new NAudio.Wave.WaveIn();
            sourceStream.WaveFormat = new NAudio.Wave.WaveFormat(16000, 1);
            waveIn = new NAudio.Wave.WaveInProvider(sourceStream);
            waveWriter = new WaveFileWriter(new IgnoreDisposeStream(memoryStream), waveIn.WaveFormat);
            sourceStream.DataAvailable += new EventHandler<NAudio.Wave.WaveInEventArgs>(sourceStream_DataAvailable);
            buff = new BufferedWaveProvider(waveIn.WaveFormat);
            sourceStream.StartRecording();
            mytimer.Enabled = true;

        }
        private void sourceStream_DataAvailable(object sender, NAudio.Wave.WaveInEventArgs e)
        {
            buff.AddSamples(e.Buffer, 0, e.BytesRecorded);

            Console.WriteLine("test");
        }
            void mytimer_Tick(object sender, EventArgs e)
        {

            if (sourceStream != null)
            {
                sourceStream.StopRecording();
                waveWriter.Flush();



                var amazonLexClient = new AmazonLexClient(Amazon.RegionEndpoint.USEast1);
                var amazonPostRequest = new Amazon.Lex.Model.PostContentRequest();
                var amazonPostResponse = new Amazon.Lex.Model.PostContentResponse();
                amazonPostRequest.BotAlias = "voiceBot";
                amazonPostRequest.BotName = "voiceBot";
                amazonPostRequest.ContentType = "audio/l16; rate=16000; channels=1";
                amazonPostRequest.UserId = "user";
                amazonPostRequest.InputStream = memoryStream;
                amazonPostRequest.UserId = "test";
                try
                {
                    amazonPostResponse = amazonLexClient.PostContent(amazonPostRequest);
                    Console.WriteLine("Got a response");
                }

                catch (Exception w)
                {
                    Console.WriteLine("{0} Exception caught.", e);
                    Console.WriteLine(w.Message);
                }

Answer 1

在将 MemoryStream 的位置传递给发布请求之前，您必须将其设置为 0。

memoryStream.Position = 0;