在Java中,如何撰写HTTP请求消息并将其发送到HTTP web服务器?


当前回答

你可以使用Socket来实现

String host = "www.yourhost.com";
Socket socket = new Socket(host, 80);
String request = "GET / HTTP/1.0\r\n\r\n";
OutputStream os = socket.getOutputStream();
os.write(request.getBytes());
os.flush();

InputStream is = socket.getInputStream();
int ch;
while( (ch=is.read())!= -1)
    System.out.print((char)ch);
socket.close();    

其他回答

我知道其他人会推荐Apache的http-客户端,但是它增加了复杂性(例如,更多可能出错的东西),这是很少被保证的。对于简单的任务,可以使用java.net.URL。

URL url = new URL("http://www.y.com/url");
InputStream is = url.openStream();
try {
  /* Now read the retrieved document from the stream. */
  ...
} finally {
  is.close();
}

你可以使用Socket来实现

String host = "www.yourhost.com";
Socket socket = new Socket(host, 80);
String request = "GET / HTTP/1.0\r\n\r\n";
OutputStream os = socket.getOutputStream();
os.write(request.getBytes());
os.flush();

InputStream is = socket.getInputStream();
int ch;
while( (ch=is.read())!= -1)
    System.out.print((char)ch);
socket.close();    

谷歌java http客户端有不错的API http请求。您可以轻松地添加JSON支持等。虽然对于简单的要求来说可能有点过分。

import com.google.api.client.http.GenericUrl;
import com.google.api.client.http.HttpRequest;
import com.google.api.client.http.HttpResponse;
import com.google.api.client.http.HttpTransport;
import com.google.api.client.http.javanet.NetHttpTransport;
import java.io.IOException;
import java.io.InputStream;

public class Network {

    static final HttpTransport HTTP_TRANSPORT = new NetHttpTransport();

    public void getRequest(String reqUrl) throws IOException {
        GenericUrl url = new GenericUrl(reqUrl);
        HttpRequest request = HTTP_TRANSPORT.createRequestFactory().buildGetRequest(url);
        HttpResponse response = request.execute();
        System.out.println(response.getStatusCode());

        InputStream is = response.getContent();
        int ch;
        while ((ch = is.read()) != -1) {
            System.out.print((char) ch);
        }
        response.disconnect();
    }
}

来自Oracle的java教程

import java.net.*;
import java.io.*;

public class URLConnectionReader {
    public static void main(String[] args) throws Exception {
        URL yahoo = new URL("http://www.yahoo.com/");
        URLConnection yc = yahoo.openConnection();
        BufferedReader in = new BufferedReader(
                                new InputStreamReader(
                                yc.getInputStream()));
        String inputLine;

        while ((inputLine = in.readLine()) != null) 
            System.out.println(inputLine);
        in.close();
    }
}

Apache HttpComponents。这两个模块的例子——HttpCore和HttpClient会让你马上开始。

并不是说HttpUrlConnection是一个糟糕的选择,HttpComponents将抽象出大量繁琐的编码。如果你真的想用最少的代码来支持大量的HTTP服务器/客户端,我推荐这样做。顺便说一下,HttpCore可以用于功能最少的应用程序(客户端或服务器),而HttpClient用于需要支持多种身份验证方案、cookie支持等的客户端。