如何在java上发送x-www-form请求？

import java.io.BufferedReader;
import java.io.DataOutputStream;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLEncoder;

public class FormRequestExample {
    public static void main(String[] args) {
        try {
            // 设置请求URL和参数
            String url = "http://example.com/api";
            String params = "param1=" + URLEncoder.encode("value1", "UTF-8") +
                            "&param2=" + URLEncoder.encode("value2", "UTF-8");

            // 创建URL对象
            URL obj = new URL(url);
            // 创建HttpURLConnection对象
            HttpURLConnection con = (HttpURLConnection) obj.openConnection();

            // 设置请求方法为POST
            con.setRequestMethod("POST");

            // 设置请求头部信息
            con.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");

            // 启用输出流
            con.setDoOutput(true);

            // 创建输出流对象
            DataOutputStream wr = new DataOutputStream(con.getOutputStream());
            // 将参数写入输出流
            wr.writeBytes(params);
            wr.flush();
            wr.close();

            // 获取响应状态码
            int responseCode = con.getResponseCode();
            System.out.println("Response Code: " + responseCode);

            // 读取响应内容
            BufferedReader in = new BufferedReader(new InputStreamReader(con.getInputStream()));
            String inputLine;
            StringBuilder response = new StringBuilder();
            while ((inputLine = in.readLine()) != null) {
                response.append(inputLine);
            }
            in.close();

            // 打印响应内容
            System.out.println("Response: " + response.toString());
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}