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java - Android TCP 客户端/服务器未正确传递消息(仅首次收到)

转载 作者:可可西里 更新时间:2023-11-01 02:52:23 25 4
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我创建了两个 android 应用程序,一个客户端和一个服务器,利用 TCP 进行跨设备通信。在服务器上,我有这段代码来监听 TCP 通信:

private class StartServer implements Runnable {     
public void run() {
getState(); // This just refreshes the server state
try {
serverSocket = new ServerSocket(5000);
appendLog("Server successfully listening ["+getLocalIpAddress() + ":5000]"); //appendLog function just uses runOnUiThread() to update the log textview
} catch (IOException e) {
appendLog("Could not listen on port: 5000");
//System.exit(-1);
}
BufferedReader in = null;
BufferedWriter out = null;
String input = null;


try {
clientSocket = serverSocket.accept();
in = new BufferedReader(new InputStreamReader(clientSocket.getInputStream()));
out = new BufferedWriter(new OutputStreamWriter(clientSocket.getOutputStream()));

while (serverState) { //variable set by our getState() function

input = in.readLine();
appendLog("Received: " + input);
this.gotMessage(out, input);
}
in.close();
out.close();
} catch (IOException e) {
//appendLog(e.toString());
appendLog("Accept failed: 5000");
//System.exit(-1);
}
}

调用此函数来实际解析收到的消息:

private void gotMessage(BufferedWriter output, String msg) {
if (msg != null) {
String sString = msg;
int to_do = 0;


if (msg.matches("^(?i)(exit|quit|close)$")) {
appendLog("Exiting");
sString = "Goodbye";
to_do=1;
} else if (msg.matches("^(?i)(launch|run|open)\\s(.+)")) {
appendLog("Launching application");
sString = "Launching application";
} else if (msg.matches("^(?i)(turn off|server off|deactivate)$")) {
appendLog("Turning off server due to remote command.");
sString = "Turning off...";
to_do=2;
} else if (msg.matches("^(?i)(restart|reboot)$")) {
appendLog("Resetting server");
sString = "Rebooting now...";
to_do=3;
}


try {
output.write("S: " + sString);
output.newLine();
output.flush();
} catch (IOException e) {
appendLog("Cannot parse message");
}


switch(to_do) {
case 0:
break;
case 1:
System.exit(0);
break;
case 2:
serverOff();
break;
case 3:
serverOff();
serverOn();
break;
default:
break;
}
}
}
}

服务器线程本身使用 Thread t = new Thread(new StartServer()); 启动; t.start()。据我所知,这很好用。我可以打开一个终端并远程登录到 IP 和端口,并且可以毫无错误地来回传递通信。但是当我尝试从下面的客户端代码做同样的事情时。我只收到第一条消息,我传入的任何其他消息都会消失在空白中。

public void sendToServer(View v) {
try {
String ip = ((TextView)findViewById(R.id.ip_box)).getText().toString(); //User entered IP address
String to_send = ((EditText)findViewById(R.id.send_to_server)).getText().toString(); //User entered text to send
this.s = new Socket(ip, 5000);
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(s.getOutputStream()));
appendLog("Sending: " + to_send);
out.write(to_send);
out.newLine();
out.close();
} catch (Exception e) {
Log.d("error", e.toString());
}
}

最佳答案

to_send 字符串中是否有行尾字符?如果不是,只需将 out.newLine() 添加到您的客户端代码中:

appendLog("Sending: " + to_send);
out.write(to_send);
out.newLine();

你的服务器代码应该像这样支持多个客户端:

// main server loop
while (serverIsActive) {
clientSocket = serverSocket.accept();
// spawn new thread for each client
ClientThread ct = new ClientThread(clientSocket);
ct.start();
}

ClientThread 应该与客户端套接字一起工作,并有自己的读取行的循环。它应该在客户端关闭套接字后立即停止:

class ClientThread {
...
public void run() {
....
while ((inputLine = in.readLine()) != null) {
// process client message
}
in.close();
}
}

关于java - Android TCP 客户端/服务器未正确传递消息(仅首次收到),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16611745/

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