我想要做的是将一些 JSON 从 Android 手机发送到 Java 服务器,它工作正常。Android/ 客户端如下所示:
Socket s = new Socket("192.168.0.36", 12390);
s.setSoTimeout(1500);
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.flush();
s.close();
服务器端是这样的:
try {
s = new ServerSocket(port);
}
catch (IOException e) {
System.out.println("Could not listen on port: " + port);
System.exit(-1);
}
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()));
String inputLine = null;
String result = "";
while ((inputLine = in.readLine()) != null) {
result = result.concat(inputLine);
}
System.out.println(result);
正如我所说,所有这些都有效。现在,我想在收到来自客户端的消息后,将消息从服务器发送回客户端。我扩展了这样的代码,Android/客户端:
Socket s = new Socket("192.168.0.36", 12390);
s.setSoTimeout(1500);
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.flush();
String inputLine = null;
String result = "";
BufferedReader in = new BufferedReader(new InputStreamReader(s.getInputStream()));
while ((inputLine = in.readLine()) != null) {
Log.d(TAG, in.readLine());
result = result.concat(inputLine);
}
和服务器端:
try {
s = new ServerSocket(port);
}
catch (IOException e) {
System.out.println("Could not listen on port: " + port);
System.exit(-1);
}
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()));
String inputLine = null;
String result = "";
while ((inputLine = in.readLine()) != null) {
result = result.concat(inputLine);
}
System.out.println(result);
PrintWriter out = new PrintWriter(c.getOutputStream());
out.write("Hello phone");
out.flush();
out.close();
在客户端,什么都没有进来,它一直挂着
while ((inputLine = in.readLine()) != null) {
Log.d(TAG, in.readLine());
result = result.concat(inputLine);
}
直到套接字超时(从不进入循环)。我认为这可能是一个时间问题,例如服务器太早发送它的回复,因此客户端从来没有收到任何东西,但我试图把 out.write("Hello phone"); 代码中的几乎任何地方,总是相同的结果。它与从 ServerSocket 获得的套接字并且无法发送数据有关吗?我在这里想念什么,这整天困扰着我...
编辑:尼古拉斯回答后,我尝试了这个(客户端):
out.write(json.toString());
out.newLine();
out.write("###");
out.flush();
String inputLine = null;
String result = "";
BufferedReader in = new BufferedReader(new InputStreamReader(s.getInputStream()));
while ((inputLine = in.readLine()) != null) {
if (inputLine.contains("###")) {
break;
}
Log.d(TAG, in.readLine());
result = result.concat(inputLine);
}
s.close();
和服务器:
while ((inputLine = in.readLine()) != null) {
result = result.concat(inputLine);
if (inputLine.contains("###")) {
System.out.println("received ###");
out.println("Hello phone");
out.println("###");
out.flush();
break;
}
}
这个想法是在客户端关闭套接字之前从服务器发送消息。仍然不起作用...任何提示?