我希望做的是将一些JSON数据从安卓手机发送到Java服务器,这个过程运行良好。安卓/客户端代码如下:
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);
如我所说,所有这些都是有效的。现在我想在从客户端接收到消息后向服务器发送一条消息。我扩展了代码,如下,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);
}
在服务器端:
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,没有任何信息进来,它会停在那里。
while ((inputLine = in.readLine()) != null) {
Log.d(TAG, in.readLine());
result = result.concat(inputLine);
}
直到套接字超时(从未进入循环)。我认为这可能是一个时间问题,例如服务器过早发送其回复,因此客户端永远不会收到任何东西,但我尝试在代码中的几乎任何地方都放置out.write("Hello phone");,结果始终相同。它是否与从ServerSocket获取套接字并无法发送数据有关?我错过了什么,这一整天都在困扰我...
编辑:在Nikolais的答案之后,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socket之前从服务器发送消息。但仍然无法正常工作...有什么提示吗?
readLine()
将不会返回。 - jtahlbornLog.d(TAG, in.readLine());
消耗掉一个额外的行 - 这会吃掉您的###
标记。 - Nikolai Fetissov