如何在 Java 中发送 HTTP 请求？ [复制]

Question

在 Java 中，如何编写 HTTP 请求消息并将其发送到 HTTP WebServer？

Answer 1

您可以使用java.net.HttpUrlConnection。

示例 (from here)，经过改进。包括在链接失效的情况下：

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();
    }
  }
}

Answer 2

来自Oracle's java tutorial

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();
    }
}

Answer 3

我知道其他人会推荐 Apache 的 http-client，但它增加了很少保证的复杂性（即，可能出错的事情更多）。对于一个简单的任务，java.net.URL 就可以了。

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();
}

Answer 4

Apache HttpComponents。这两个模块的示例 - HttpCore 和 HttpClient 将帮助您立即开始。

并不是说 HttpUrlConnection 是一个糟糕的选择，HttpComponents 会抽象掉很多繁琐的编码。如果你真的想用最少的代码支持很多 HTTP 服务器/客户端，我会推荐这个。顺便说一句，HttpCore 可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。

Answer 5

有一个关于通过 Example Depot 发送 POST 请求 here 的很棒的链接::

try {
    // Construct data
    String data = URLEncoder.encode("key1", "UTF-8") + "=" + URLEncoder.encode("value1", "UTF-8");
    data += "&" + URLEncoder.encode("key2", "UTF-8") + "=" + URLEncoder.encode("value2", "UTF-8");

    // Send data
    URL url = new URL("http://hostname:80/cgi");
    URLConnection conn = url.openConnection();
    conn.setDoOutput(true);
    OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
    wr.write(data);
    wr.flush();

    // Get the response
    BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
    String line;
    while ((line = rd.readLine()) != null) {
        // Process line...
    }
    wr.close();
    rd.close();
} catch (Exception e) {
}

如果您想发送 GET 请求，您可以稍微修改代码以满足您的需要。具体来说，您必须在 URL 的构造函数中添加参数。然后，也注释掉这个wr.write(data);

你应该注意的一点没有写出来，那就是超时。特别是如果你想在 WebServices 中使用它，你必须设置超时，否则上面的代码将无限期地等待或至少等待很长时间，这可能是你不想要的。

超时设置如下conn.setReadTimeout(2000);输入参数以毫秒为单位

Answer 6

这将对您有所帮助。不要忘记将 JAR HttpClient.jar 添加到类路径中。

import java.io.FileOutputStream;
import java.io.IOException;

import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.HttpStatus;
import org.apache.commons.httpclient.NameValuePair;
import org.apache.commons.httpclient.methods.PostMethod;

public class MainSendRequest {

     static String url =
         "http://localhost:8080/HttpRequestSample/RequestSend.jsp";

    public static void main(String[] args) {

        //Instantiate an HttpClient
        HttpClient client = new HttpClient();

        //Instantiate a GET HTTP method
        PostMethod method = new PostMethod(url);
        method.setRequestHeader("Content-type",
                "text/xml; charset=ISO-8859-1");

        //Define name-value pairs to set into the QueryString
        NameValuePair nvp1= new NameValuePair("firstName","fname");
        NameValuePair nvp2= new NameValuePair("lastName","lname");
        NameValuePair nvp3= new NameValuePair("email","email@email.com");

        method.setQueryString(new NameValuePair[]{nvp1,nvp2,nvp3});

        try{
            int statusCode = client.executeMethod(method);

            System.out.println("Status Code = "+statusCode);
            System.out.println("QueryString>>> "+method.getQueryString());
            System.out.println("Status Text>>>"
                  +HttpStatus.getStatusText(statusCode));

            //Get data as a String
            System.out.println(method.getResponseBodyAsString());

            //OR as a byte array
            byte [] res  = method.getResponseBody();

            //write to file
            FileOutputStream fos= new FileOutputStream("donepage.html");
            fos.write(res);

            //release connection
            method.releaseConnection();
        }
        catch(IOException e) {
            e.printStackTrace();
        }
    }
}

Answer 7

你可以像这样使用 Socket

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();    

Answer 8

这是一个完整的 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://tools.ietf.org/rfc/rfc768.txt").openStream())) {
      System.out.println(s.useDelimiter("\\A").next());
    }
  }
}

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

Answer 9

Google java http client 为 http 请求提供了很好的 API。您可以轻松地添加 JSON 支持等。虽然对于简单的请求，它可能是矫枉过正。

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();
    }
}