2012-07-26 84 views
1

我试图让(android)客户端与服务器进行通信。发送数据客户端 - >服务器工作正常,但我希望服务器响应。客户端代码:Java服务器 - 客户端套接字通信

try { 
       Socket s = new Socket("192.168.0.36", 12390); 
       s.setSoTimeout(5000); 

       JSONObject json = new JSONObject(); 
       json.put("emergency", false); 
       json.put("imei", imei); 
       json.put("lat", l.getLatitude()); 
       json.put("lon", l.getLongitude()); 
       json.put("acc", l.getAccuracy()); 
       json.put("time", l.getTime()); 


       BufferedWriter out = new BufferedWriter(new OutputStreamWriter(
       s.getOutputStream())); 
       out.write(json.toString()); 
       out.newLine(); 
       out.write("###"); 
       out.flush(); 
       Log.d(TAG, "sent"); 


       String inputLine = null; 
       String result = ""; 
       BufferedReader in = new BufferedReader(new InputStreamReader(s.getInputStream())); 

       Log.d(TAG, "open input"); 
       while ((inputLine = in.readLine()) != null) { 
        Log.d(TAG, "while"); 
        if (inputLine.contains("###")) { 
         break; 
        } 
        result = result.concat(inputLine); 

       } 

       Log.d(TAG, "closing socket"); 
       s.close(); 

服务器端:

Socket c = null; 
    while (true) { 
     try { 
      c = s.accept(); 
     } catch (IOException e) { 
      System.out.println("Accept failed: " + port); 
      System.exit(-1); 
     } 
     try { 

      BufferedReader in = new BufferedReader(new InputStreamReader(c.getInputStream())); 
      BufferedWriter out = new BufferedWriter(new OutputStreamWriter(c.getOutputStream())); 



      String inputLine = null; 
      String result = ""; 
      while ((inputLine = in.readLine()) != null) { 

       if (inputLine.contains("###")) { 
        System.out.println("received ###"); 
         out.write("Hello phone"); 
         out.newLine(); 
         out.write("###"); 
         out.newLine(); 
         out.flush(); 

       } 

       result = result.concat(inputLine); 


      } 
      System.out.println(result); 

      out.close(); 

服务器正确读取来自客户端的消息和接收###应发回的消息后,却是从未收到由客户。客户端超时

    while ((inputLine = in.readLine()) != null) { 
        Log.d(TAG, "while"); 
        if (inputLine.contains("###")) { 
         break; 
        } 

永远不会进入while循环。我在这里错过了什么?任何想法赞赏!

回答

2

从客户端:

out.write(json.toString()); 
out.newLine(); 
out.write("###"); 
out.flush(); 

你忘了添加后发送 “###” 新行,因为你的服务器使用的readLine()

while ((inputLine = in.readLine()) != null) { 
//... 
} 

所以,我相信您的服务器可以不会收到此消息。尝试把

out.newLine(); 

发送“###”后。

+0

这实际上是它!我没有注意到它,因为服务器记录了“System.out.println(”received ###“);”,但那个人解决了它。谢谢一堆! – FWeigl 2012-07-26 09:46:10

0

我可能在这里是非常错误的,但根据我的经验,我总是在我开始通信之前打开进出流。请尝试移动

BufferedReader in = new BufferedReader(new InputStreamReader(s.getInputStream())); 

旁边

BufferedWriter out = new BufferedWriter(new OutputStreamWriter(s.getOutputStream())); 

万一你打开你的输入流,并开始阅读之前你的数据包过期。

+0

试过了,没有帮助,但感谢您的输入! – FWeigl 2012-07-26 09:00:51