GET request with Java sockets

2020-07-13 11:38发布

I am writing a simple program to send a get request to a specific url "http://badunetworks.com/about/". The request works if I send it to "http://badunetworks.com" but I need to send it to the about page.

package badunetworks;
import java.io.*;
import java.net.*;

public class GetRequest {


    public static void main(String[] args) throws Exception {

        GetRequest getReq = new GetRequest();

        //Runs SendReq passing in the url and port from the command line
        getReq.SendReq("www.badunetworks.com/about/", 80);


    }

    public void SendReq(String url, int port) throws Exception {

        //Instantiate a new socket
        Socket s = new Socket("www.badunetworks.com/about/", port);

        //Instantiates a new PrintWriter passing in the sockets output stream
        PrintWriter wtr = new PrintWriter(s.getOutputStream());

        //Prints the request string to the output stream
        wtr.println("GET / HTTP/1.1");
        wtr.println("Host: www.badunetworks.com");
        wtr.println("");
        wtr.flush();

        //Creates a BufferedReader that contains the server response
        BufferedReader bufRead = new BufferedReader(new InputStreamReader(s.getInputStream()));
        String outStr;

        //Prints each line of the response 
        while((outStr = bufRead.readLine()) != null){
            System.out.println(outStr);
        }


        //Closes out buffer and writer
        bufRead.close();
        wtr.close();

    }

}

3条回答
Deceive 欺骗
2楼-- · 2020-07-13 11:56

if the about page link is about.html ,then you have change this line wtr.println("GET / HTTP/1.1") into wtr.println("GET /about.html HTTP/1.1").

in socket creation remove the /about

wtr.println("GET / HTTP/1.1");--->this line call the home page of the host you specified.

查看更多
该账号已被封号
3楼-- · 2020-07-13 12:00

you need open Socket to url without path e.g.

Socket("www.badunetworks.com", port);

and after send command GET /{path} HTTP/1.1 e.g.

GET /about HTTP/1.1

... other header...

查看更多
手持菜刀,她持情操
4楼-- · 2020-07-13 12:02

When you're doing such a low-level access to a webserver, you should understand the 7 OSI layers. Socket is on layer 5 and HTTP on layer 7. That's also the reason, why java.net.Socket only accepts host names or InetAddr and no URLs. To do it with sockets, you have to implement the HTTP protocol properly, that is

  • Create a socket connection to the host and port, i.e. www.badunetworks.com and 80
  • Send a HTTP packet to the outputstream containing, method, resource by path and protocol version, i.e. GET /about/ HTTP/1.1
  • Read and interpret the response properly (headers and body)

But I wonder why you're doing it this complicated, there are a lot of alternatives to implementing low-level http clients yourself:

  • good old java.net.URL, as deprecated as its handling is, it's still one of the easiest way to read a resource, just call openStream() to read it
  • Apache HTTP Client is one of the most widely used http client implementation for java, which is quite easy to use and more flexible than the reading via URL
  • javax.ws.rs has a good builder api for creating web clients
查看更多
登录 后发表回答