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

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();
    }
}

其他回答

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();
    }
}

您可以使用java.net.HttpUrlConnection。

示例(来自这里),并进行了改进。包括在链接腐烂的情况下:

public static String executePost(String targetURL, String urlParameters) {
  HttpURLConnection connection = null;

  try {
    //Create connection
    URL url = new URL(targetURL);
    connection = (HttpURLConnection) url.openConnection();
    connection.setRequestMethod("POST");
    connection.setRequestProperty("Content-Type", 
        "application/x-www-form-urlencoded");

    connection.setRequestProperty("Content-Length", 
        Integer.toString(urlParameters.getBytes().length));
    connection.setRequestProperty("Content-Language", "en-US");  

    connection.setUseCaches(false);
    connection.setDoOutput(true);

    //Send request
    DataOutputStream wr = new DataOutputStream (
        connection.getOutputStream());
    wr.writeBytes(urlParameters);
    wr.close();

    //Get Response  
    InputStream is = connection.getInputStream();
    BufferedReader rd = new BufferedReader(new InputStreamReader(is));
    StringBuilder response = new StringBuilder(); // or StringBuffer if Java version 5+
    String line;
    while ((line = rd.readLine()) != null) {
      response.append(line);
      response.append('\r');
    }
    rd.close();
    return response.toString();
  } catch (Exception e) {
    e.printStackTrace();
    return null;
  } finally {
    if (connection != null) {
      connection.disconnect();
    }
  }
}

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。

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();
    }
}

下面是一个完整的Java 7程序:

class GETHTTPResource {
  public static void main(String[] args) throws Exception {
    try (java.util.Scanner s = new java.util.Scanner(new java.net.URL("http://example.com/").openStream())) {
      System.out.println(s.useDelimiter("\\A").next());
    }
  }
}

新的try-with-resources将自动关闭Scanner,而Scanner将自动关闭InputStream。