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我已经实现了一个非常非常简单的Java服务器,它基本上监听端口8080上的所有请求并将它们写入控制台。Android - Java服务器和Android客户端
我的代码工作正常,一次。但是,对于每次重新启动应用程序,服务器只会收到一个请求。 任何人都可以帮助我吗?
Server代码:
public class RunServer {
public static void main(String[] args) {
final int port = 8080;
final String path = "/android";
try {
HttpServer server = HttpServer.create(new InetSocketAddress(port), 0); //Create the server to listen for incoming requests on the specified port
server.createContext(path, new HttpHandler() { //Set a handler for incoming requests to the specific path
@Override
public void handle(HttpExchange arg0) throws IOException { //This method is called when new requests are incoming
InputStreamReader isr = new InputStreamReader(arg0.getRequestBody(),"utf-8");
BufferedReader br = new BufferedReader(isr);
String query = br.readLine();
System.out.print(query);
} });
server.setExecutor(null); //Default executor. Other executors may be used to e.g. use a threadpool for handling concurrent incoming requests.
server.start(); //Start the server.
} catch (Exception e) {
e.printStackTrace();
}
}
}
我使用这个客户端发布到我的服务器:
public class MyAsyncTask extends AsyncTask<String, String, Void>{
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://192.168.1.5:8080/android");
String s;
@Override
protected Void doInBackground(String... params) {
// Building post parameters
// key and value pair
List<NameValuePair> nameValuePair = new ArrayList<NameValuePair>(2);
nameValuePair.add(new BasicNameValuePair("email", "[email protected]"));
nameValuePair.add(new BasicNameValuePair("message",
"Hi, trying Android HTTP post!"));
// Url Encoding the POST parameters
try {
httpPost.setEntity(new UrlEncodedFormEntity(nameValuePair));
} catch (UnsupportedEncodingException e) {
// writing error to Log
e.printStackTrace();
}
// Making HTTP Request
try {
HttpResponse response = httpClient.execute(httpPost);
// writing response to log
Log.d("Http Response:", response.toString());
} catch (ClientProtocolException e) {
// writing exception to log
e.printStackTrace();
} catch (IOException e) {
// writing exception to log
e.printStackTrace();
}
return null;
}
}
谢谢,我认为你对第一部分是正确的。我试图用tomcat设置它作为一个servlet,它现在看起来很完美! –
如果您最终将其用于生产,请务必阅读文档的安全部分。如果你有一个现有的Java应用程序,也许考虑嵌入Jetty或Tomcat而不是独立。 – Drizzt321