请给出非常简单的android studio的POST命令代码

import android.os.AsyncTask;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.net.HttpURLConnection;
import java.net.URL;

public class PostRequestTask extends AsyncTask<String, Void, String> {

    @Override
    protected String doInBackground(String... params) {
        String urlString = params[0];
        String postData = params[1];
        String response = "";

        try {
            URL url = new URL(urlString);
            HttpURLConnection connection = (HttpURLConnection) url.openConnection();
            connection.setRequestMethod("POST");
            connection.setRequestProperty("Content-Type", "application/json");
            connection.setDoOutput(true);

            OutputStream outputStream = connection.getOutputStream();
            outputStream.write(postData.getBytes());
            outputStream.flush();
            outputStream.close();

            int statusCode = connection.getResponseCode();
            if (statusCode == HttpURLConnection.HTTP_OK) {
                BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
                String line;
                while ((line = reader.readLine()) != null) {
                    response += line;
                }
                reader.close();
            } else {
                response = "Error: " + statusCode;
            }

            connection.disconnect();
        } catch (IOException e) {
            e.printStackTrace();
            response = "Error: " + e.getMessage();
        }

        return response;
    }

    @Override
    protected void onPostExecute(String result) {
        // 处理POST请求的响应结果
    }
}

String url = "https://example.com/api";
String postData = "{\"key\": \"value\"}";

PostRequestTask task = new PostRequestTask();
task.execute(url, postData);