在正确的查询字符串中附加到android中的改造请求

import java.io.IOException;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLEncoder;
import java.nio.charset.StandardCharsets;

public class QueryStringExample {
    public static void main(String[] args) {
        try {
            String baseUrl = "https://example.com/search";
            String queryParam = "q=" + URLEncoder.encode("android", StandardCharsets.UTF_8.toString());
            String langParam = "lang=" + URLEncoder.encode("en", StandardCharsets.UTF_8.toString());
            String urlString = baseUrl + "?" + queryParam + "&" + langParam;

            URL url = new URL(urlString);
            HttpURLConnection connection = (HttpURLConnection) url.openConnection();
            connection.setRequestMethod("GET");

            int responseCode = connection.getResponseCode();
            System.out.println("Response Code: " + responseCode);

            // 处理响应
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}