如何在安卓上进行异步URL连接？

Question

我正在使用以下类连接到我的 Web 服务。我想使它异步化。我该怎么做？

package org.stocktwits.helper;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;

import android.util.Log;

public class RestClient{

    private static String convertStreamToString(InputStream is) {
        /*
         * To convert the InputStream to String we use the BufferedReader.readLine()
         * method. We iterate until the BufferedReader return null which means
         * there's no more data to read. Each line will appended to a StringBuilder
         * and returned as String.
         */
        BufferedReader reader = new BufferedReader(new InputStreamReader(is));
        StringBuilder sb = new StringBuilder();

        String line = null;
        try {
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                is.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

        return sb.toString();
    }


    /* This is a test function which will connects to a given
     * rest service and prints it's response to Android Log with
     * labels "Praeda".
     */
    public static JSONObject connect(String url)
    {

        HttpClient httpclient = new DefaultHttpClient();


        // Prepare a request object
        HttpGet httpget = new HttpGet(url); 

        // Execute the request
        HttpResponse response;
        try {
            response = httpclient.execute(httpget);
            // Examine the response status
            Log.i("Praeda",response.getStatusLine().toString());

            // Get hold of the response entity
            HttpEntity entity = response.getEntity();

            if (entity != null) {

                // A Simple JSON Response Read
                InputStream instream = entity.getContent();
                String result= convertStreamToString(instream);

                // A Simple JSONObject Creation
                JSONObject json=new JSONObject(result);

                // Closing the input stream will trigger connection release
                instream.close();

                return json;
            }


        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

        return null;
    }

}

- Sheehan Alam

---

请参考以下链接：https://dev59.com/c3NA5IYBdhLWcg3wUL_n - Ladlestein

Answer 1

除了Ladlestein评论中提到的所有可能解决方案外，还有一个简单的答案，就是将所有内容包装在一个AsyncTask中。

- CommonsWare

---

嗨。我知道这是一个非常老的问题，因此是一个旧答案，但您是否介意更新示例项目以在Android框架中进行异步HTTP请求？谢谢！ - Con Antonakos

Answer 2

你最好在github上使用httpRequest。这是链接给你： http://kevinsawicki.github.io/http-request/

- Alexander.Iljushkin