HTTPUrlConnection错误（从inputStream中读取后无法打开OutputStream）

Question

我是Java的新手，在使用HTTPURLConnection在Android上发送多个发布请求时遇到上述错误。我已经写了一个HTTPTransport类，其中我想要sendMessage和recvMessage方法。

public class HTTPTransport 
{ 
    private HttpURLConnection connection; 

    public HTTPTransport() 
    { 
     URL url = new URL("http://test.com"); 

     connection = (HttpURLConnection) url.openConnection(); 
     connection.setRequestMethod("POST"); 
     connection.setDoInput(true); 
     connection.setDoOutput(true); 
     connection.setRequestProperty("Content-Type", "application/octet-stream"); 
     connection.setRequestProperty("Accept-Encoding", "gzip"); 
     connection.setRequestProperty("Connection", "Keep-Alive"); 
    } 

    public void sendMessage(byte[] msgBuffer, long size) 
    { 
     try 
     { 
     DataOutputStream dos = new DataOutputStream(connection.getOutputStream()); 
     dos.write(msgBuffer, 0, (int)size); 
     dos.flush(); 
     dos.close(); 

     dos.close(); 
     } 
     catch(IOException e) 
     { 
     // This exception gets triggered with the message mentioned in the title. 
     Log.e(TAG, "IOException: " + e.toString()); 
     } 
    } 
    public byte[] recvMessage() 
    { 

     int readBufLen = 1024; 

     byte[] buffer = new byte[readBufLen]; 

     int len = 0; 
     FileOutputStream fos = new FileOutputStream(new File("/sdcard/output.raw")); 

     DataInputStream dis = new DataInputStream(connection.getInputStream()); 
     while((len = dis.read(buffer, 0, readBufLen)) > 0) 
     { 
     Log.d(TAG, "Len of recd bytes " + len + ", Byte 0 = " + buffer[0]); 
     //Save response to a file 
     fos.write(buffer, 0, len); 
     } 

     fos.close(); 
     dis.close(); 
     return RecdMessage;  
    } 
} 

我能够使用sendMessage和recvMessage成功发送第一条消息。当我尝试发送第二个错误时，我看到以下错误： IOException：java.net.ProtocolException：从输入流读取后无法打开OutputStream

请让我知道如何编写此类。

谢谢！

Answer 1

您的执行HTTPUrlConnectiondoes not allow you to reuse the connection in this manner。我相信您必须使用HttpConnectionManager才能以您想要的方式使用Keep-Alive。

Answer 2

您需要为每个请求使用新的HttpURLConnection。 TCP连接本身将集中在幕后。不要试图自己做。