我想在java中编写一个简单的http服务器。这里是我到目前为止的代码:Java套接字写入然后从套接字读取
服务器:
import java.net.*;
import java.io.*;
import java.util.*;
import java.util.regex.*;
public class Server
{
static final int PORT = 8080;
final String REQUEST_FORMAT = "^GET (.*?) HTTP/1.1$";
final Socket client;
public Server(Socket s)
{
client = s;
}
public void run()
{
try
(Scanner in = new Scanner(new InputStreamReader(client.getInputStream()));
PrintWriter out = new PrintWriter(client.getOutputStream(),true);)
{
String request = in.findInLine(Pattern.compile(REQUEST_FORMAT));
System.out.println(request);
out.write("HTTP/1.1 200 OK");
}
catch(Exception ex)
{
ex.printStackTrace();
}
}
public static void main(String []args)
{
try
(
ServerSocket server = new ServerSocket(PORT);
Socket client = server.accept();
)
{
Server s = new Server(client);
s.run();
}
catch(Exception ex)
{
ex.printStackTrace();
}
}
}
客户:
import java.net.*;
import java.io.*;
import java.util.*;
public class Client
{
final int PORT = 8080;
String data = "GET /home/index.html HTTP/1.1";
public Client()
{
try
(Socket socket = new Socket("127.0.0.1",PORT);
PrintWriter out = new PrintWriter(socket.getOutputStream(),true);
Scanner in = new Scanner(new InputStreamReader(socket.getInputStream()));)
{
out.print(data);
String header1 = in.next();
System.out.println("header="+header1);
int status = in.nextInt();
System.out.println("status="+status);
String message = in.next();
System.out.println("message="+message);
}
catch(Exception ex)
{
}
}
public static void main(String []args)
{
Client c = new Client();
}
}
目前客户端只是写了一个样本请求到服务器,服务器写入的样本响应头。但是,服务器似乎在读取请求之后无限期地等待客户端输入,而无需继续发送响应。请帮助解决这个问题。在
String request = in.findInLine(Pattern.compile(REQUEST_FORMAT));
但是,如果我在最后写新行,这不是错的。据我所知,HTTP GET请求最后没有换行符? – PPGoodMan 2015-02-23 17:51:37
@PP好心人你错了。不要做出无证据的猜测。 – EJP 2015-02-23 18:01:38
您是否试图实现一个简单的套接字服务器和套接字客户端?它有用吗?从那里你可以做出改进,使它看起来像一个Http套接字服务器 – alainlompo 2015-02-23 18:43:35