java - 如何在Java中发送HTTP请求？ [重复](How to send HTTP request in java? [duplicate])

Question

This question already has an answer here:

(这个问题已经在这里有了答案：)

• How to use java.net.URLConnection to fire and handle HTTP requests 11 answers

  (如何使用java.net.URLConnection触发和处理HTTP请求 11个答案)

In Java, How to compose a HTTP request message and send it to a HTTP WebServer?

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

  ask by Yatendra Goel translate from so

Answer 1

You can use java.net.HttpUrlConnection .

(您可以使用java.net.HttpUrlConnection 。)

Example ( from here ), with improvements.

(示例（ 从此处开始 ），进行了改进。)

Included in case of link rot:

(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('
');
    }
    rd.close();
    return response.toString();
  } catch (Exception e) {
    e.printStackTrace();
    return null;
  } finally {
    if (connection != null) {
      connection.disconnect();
    }
  }
}