在 Android 上从 URL 检索 JSON

Question

我的手机APP以文本模式完美下载内容。下面是执行此操作的代码。我调用 Communicator 类并执行 HttpGet：

URL_Data = new Communicator().executeHttpGet("Some URL");

public class Communicator {
public String executeHttpGet(String URL) throws Exception 
{
    BufferedReader in = null;
    try 
    {
        HttpClient client = new DefaultHttpClient();
        client.getParams().setParameter(CoreProtocolPNames.USER_AGENT, "android");
        HttpGet request = new HttpGet();
        request.setHeader("Content-Type", "text/plain; charset=utf-8");
        request.setURI(new URI(URL));
        HttpResponse response = client.execute(request);
        in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));

        StringBuffer sb = new StringBuffer("");
        String line = "";

        String NL = System.getProperty("line.separator");
        while ((line = in.readLine()) != null) 
        {
            sb.append(line + NL);
        }
        in.close();
        String page = sb.toString();
        //System.out.println(page);
        return page;
    } 
    finally 
    {
        if (in != null) 
        {
            try 
            {
                in.close();
            } 
            catch (IOException e)    
            {
                Log.d("BBB", e.toString());
            }
        }
    }
}
}

我收到的是这个（网址的源代码）：

[{"id_country":"3","country":"AAA"},{"id_country":"8","country":"BBB"},
{"id_country":"66","country":"CCC"},{"id_country":"14","country":"DDD"},
{"id_country":"16","country":"EEE"},{"id_country":"19","country":"FFF"},
{"id_country":"24","country":"GGG"},{"id_country":"33","country":"HHH"},
{"id_country":"39","country":"III"},{"id_country":"44","country":"JJJ"},
{"id_country":"45","country":"KKK"},{"id_country":"51","country":"LLL"},
{"id_country":"54","country":"MMM"},{"id_country":"55","country":"NNN"},
{"id_country":"57","country":"OOO"},{"id_country":"58","country":"PPP"},
{"id_country":"63","country":"RRR"},{"id_country":"65","country":"SSS"}]

这个响应是一个字符串。在服务器上，它作为 JSON 对象输出（使用 PHP），现在在我的 Android PHP 中，我想将此字符串转换为 JSON。这可能吗？

Answer 1

您收到的是来自InputStream 的一系列字符，您附加到StringBuffer 并在最后转换为String - 所以String 的结果是可以的:)

你想要的是通过org.json.* 类对这个字符串进行后处理

String page = new Communicator().executeHttpGet("Some URL");
JSONObject jsonObject = new JSONObject(page);

然后处理jsonObject。由于你收到的数据是一个数组，你其实可以说

String page = new Communicator().executeHttpGet("Some URL");
JSONArray jsonArray = new JSONArray(page);
for (int i = 0 ; i < jsonArray.length(); i++ ) {
  JSONObject entry = jsonArray.get(i);
  // now get the data from each entry
}

Answer 2

编辑：

要进一步了解我之前链接到您的问题，请使用此示例。把它放在一个返回 JSONArray 的函数中（这样你就可以遍历数组并使用 array.getString）。这适用于大多数数据量。它会将正确的压缩头发送到 Web 服务器并检测 gzip 压缩结果。试试看：

    URL url = new URL('insert your uri here');
    HttpURLConnection urlConn = (HttpURLConnection) url.openConnection();
    urlConn.setRequestProperty("Accept-Encoding", "gzip");
    HttpURLConnection httpConn = (HttpURLConnection) urlConn;
    httpConn.setAllowUserInteraction(false);
    httpConn.connect();
    if (httpConn.getContentEncoding() != null) {
        String contentEncoding = httpConn.getContentEncoding().toString();
        if (contentEncoding.contains("gzip")) {
        in = new GZIPInputStream(httpConn.getInputStream());
        }
        // else it is encoded and we do not want to use it...
    } else {
        in = httpConn.getInputStream();
    }
    BufferedInputStream bis = new BufferedInputStream(in);
    ByteArrayBuffer baf = new ByteArrayBuffer(1000);
    int read = 0;
    int bufSize = 1024;
    byte[] buffer = new byte[bufSize];
    while (true) {
        read = bis.read(buffer);
        if (read == -1) {
        break;
        }
        baf.append(buffer, 0, read);
    }
    queryResult = new String(baf.toByteArray());
    return new JSONArray(queryResult);

/结束编辑

尝试阅读我在这个 SO 问题上发布的解决方案：

Create list in android app

第，

学习

Answer 3

使用 org.json.JSONObject

JSONObject json = new JSONObject(oage);

需要注意的是响应只是“真”或“假”。可能想要创建一个 util 函数来检查这些情况，否则只需加载 JSONObject。

好的，在这种情况下你会使用 JSONArray

JSONArray jsonArray = new JSONArray(page); 
for (int i = 0; i < jsonArray.length(); ++i) {
  JSONObject element = jsonArray.getJSONObject(i);
  ..... 
}

Answer 4

您是否尝试将内容类型设置为application/json？

Answer 5

假设我们有一个名为 Post

的 POJO 类

public List<Post> getData(String URL) throws InterruptedException, ExecutionException {
    //This has to be AsyncTask because data streaming from remote web server should be running in background thread instead of main thread. Or otherwise your application will hung while connecting and getting data.
    AsyncTask<String,String, List<Post>> getTask = new AsyncTask<String,String,List<Post>>(){
        @Override
        protected List<Post> doInBackground(String... params) {
            List<Post> postList  = new ArrayList<Post>();
            String response = "";
            try{
                //Read stream data from url START
                java.net.URL url = new java.net.URL(params[0]);
                HttpURLConnection urlConnection = (HttpURLConnection)
                        url.openConnection();
                BufferedReader reader = new  BufferedReader(new InputStreamReader(urlConnection.getInputStream()));
                String line = "";
                while((line = reader.readLine()) != null){
                    response += line + "\n";
                }
                //Read stream data from url END

                //Parsing json data from reponse data START
                JSONArray jsonArray = new JSONArray(response);
                for(int i=0;i<jsonArray.length();i++){

                    String message = jsonArray.getJSONObject(i).getString("message");
                    // Post class has a constructor which accept message value.
                    postList.add(new Post(message));
                }
                //Parsing json data from reponse data END
            } catch (Exception e){
                e.printStackTrace();
            }

            return postList;
        }

        protected void onPostExecute(String result) {
        }
    };
    //This will return a list of posts
    return getTask.execute(URL).get();
}

Answer 6

试试就好

///...

JSONArray jsonArray = new JSONArray(responseString);

for(JSONObject jsonObject:jsonArray) { .........
}

////..