Question

我正在尝试手动构建HTTP请求（作为字符串）并通过TCP套接字发送，这就是我要做的事情：

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.net.MalformedURLException;
import java.net.Socket;
import java.net.URL;
import java.util.logging.Level;
import java.util.logging.Logger;

/**
 * SimpleHttpClient.java (UTF-8)
 *
 * Mar 27, 2014
 *
 * @author tarrsalah.org
 */
public class SimpleHttpClient {
    private static final String StackOverflow = "http://stackoverflow.com/";

    public static void main(String[] args) {

//      if (args.length < 1) {
//          System.out.println("Usage : SimpleHttpClient <url>");
//          return;
//      }

        try {
            URL url = new URL(StackOverflow);
            String host = url.getHost();
            String path = url.getPath();
            int port = url.getPort();
            if (port < 80) {
                port = 80;
            }

            //Construct and send the HTTP request
            String request = "GET" + path + "HTTP/1.1\n";
            request += "host: " + host;
            request += "\n\n";

            // Open a TCP connection
            Socket socket  = new Socket(host, port);
            // Send the request over the socket
            PrintWriter writer = new PrintWriter(socket.getOutputStream());
            writer.print(request);
            writer.flush();
            // Read the response
            BufferedReader reader = new BufferedReader(new InputStreamReader(socket.getInputStream()));
            String next_record = null;
            while ((next_record = reader.readLine()) != null) {
                System.out.println(next_record);
            }
            socket.close();
        } catch (MalformedURLException ex) {
            Logger.getLogger(SimpleHttpClient.class.getName()).log(Level.SEVERE, null, ex);
        } catch (IOException ex) {
            Logger.getLogger(SimpleHttpClient.class.getName()).log(Level.SEVERE, null, ex);
        }
    }
}

但无论选择哪个网址，我都会收到此回复消息：

HTTP/1.0 400 Bad request
Cache-Control: no-cache
Connection: close
Content-Type: text/html

<html><body><h1>400 Bad request</h1>
Your browser sent an invalid request.
</body></html>

为什么？

Answer 1

感谢Julian Reschke的建议，我错过了两个空格字符（一个在HTTP动词之后，第二个在路径之后）

String request = "GET " + path + " HTTP/1.1\n";
//                   ^            ^

Answer 2

我实际上有点不确定这是否能解决您的问题，但是如果您阅读RFC2616, section 5，您会注意到CRLF之后Request-line非常明确地提及\r和标题，因此您可能会错过代码中的某些{{1}}。

干杯，

为什么在通过TCP套接字发送手动制作的HTTP请求时，我收到HTTP 400错误的请求响应？

2 个答案: