使用 android 上的 HTTPURL 连接从 URL 获取 JSON 对象

Question

我目前正在尝试访问我的 android 应用程序上的 JSON 对象。但是，我不断收到以下错误：

android.os.NetworkOnMainThreadException

通过稍微研究一下，我目前唯一的信息是我需要异步执行此操作(?)

这是我的代码：

public class MainActivity extends AppCompatActivity {

@Override
protected void onCreate(Bundle savedInstanceState) {

    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    TextView dynamictext;
    dynamictext = (TextView) findViewById(R.id.dynamictext);
    dynamictext.setText(getJSON("my url here"));

}


public String getJSON(String url) {
    HttpURLConnection c = null;
    try {
        URL u = new URL(url);
        c = (HttpURLConnection) u.openConnection();
        c.connect();
        int status = c.getResponseCode();
        switch (status) {
            case 200:
            case 201:
                BufferedReader br = new BufferedReader(new InputStreamReader(c.getInputStream()));
                StringBuilder sb = new StringBuilder();
                String line;
                while ((line = br.readLine()) != null) {
                    sb.append(line+"\n");
                }
                br.close();
                return sb.toString();
        }

    } catch (Exception ex) {
        return ex.toString();
    } finally {
        if (c != null) {
            try {
                c.disconnect();
            } catch (Exception ex) {
                //disconnect error
            }
        }
    }
    return null;
}
}

提前致谢

Answer 1

从 Android 3 及更高版本开始，不允许在 UI 线程上进行网络操作运行，因此您的应用程序将因 NetworkOnMainThreadException 而崩溃。

在这里您可以找到一个 Asyntask 的简单示例，您将异步处理您的代码：

Answer 2

我推荐阅读这篇文章url

添加异步任务；

public class TestAsyncTask extends AsyncTask<Void, Void, String> {
    private Context mContext;
    private String mUrl;

    public TestAsyncTask(Context context, String url) {
        mContext = context;
        mUrl = url;
    }

    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        dynamictext = (TextView) findViewById(R.id.dynamictext);
    }

    @Override
    protected String doInBackground(Void... params) {
        String resultString = null;
        resultString = getJSON(mUrl);

        return resultString;
    }

    @Override
    protected void onPostExecute(String strings) {
        super.onPostExecute(strings);
        dynamictext.setText(strings);
    }

    private String getJSON(String url) { ... }
}

使用异步任务；

public class MainActivity ... {
    private TextView dynamictest;

    onCreate() {
        ...
        TestAsyncTask testAsyncTask = new TestAsyncTask(MainActivity.this, "my url here");
        testAsyncTask.execute();
    }
}

Answer 3

要执行耗时的任务，您需要实施异步任务，该任务将在非 UI 线程上运行，因为您不能在 UI 线程上执行此类操作。

并且您可以在 activity 的 onStart() 方法中执行异步任务，并覆盖异步任务的 onPostMethod() 以实现您希望对 UI 执行的任何更改一旦网络运行检查 http://developer.android.com/reference/android/os/AsyncTask.html 以了解有关异步任务的详细信息。

Answer 4

private void taskPostExecute(String url)
{
     myTask2=(RequestTask)new RequestTask().execute(url);
}

class RequestTask extends AsyncTask<String, String, String> {
    @Override
    protected String doInBackground(String... uri) {
        try {
                postData(uri[0]);
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        }

        return null;
    }
    @Override
    protected void onCancelled() {
        myTask2.cancel(true);

    }
    @Override
    protected void onPostExecute(String url) {
        LinearLayouts();
    }
}

public static String postData(String url) throws UnsupportedEncodingException {
    HttpClient httpclient = new DefaultHttpClient();
    HttpGet httppost = new HttpGet(url);
    String responseString = "";

    try {
        HttpResponse response = httpclient.execute(httppost);
        StatusLine statusLine = response.getStatusLine();
        InputStream inputStream = null;
        if(statusLine.getStatusCode() == HttpStatus.SC_OK){
            inputStream = response.getEntity().getContent();
            if(inputStream != null)
                responseString = convertInputStreamToString(inputStream);
            else
                responseString = "Did not work!";
            JSONObject reader = null;
            JSONArray sys  = null;
            try {
                reader = new JSONObject(responseString);
                String res_metar = reader.getString("TAG");

            } catch (JSONException e) {
                e.printStackTrace();
            }
            //a = responseString;

        } else{
            response.getEntity().getContent().close();
            throw new IOException(statusLine.getReasonPhrase());
        }
    } catch (ClientProtocolException e) {

    } catch (IOException e) {

        // process execption
    }
    return responseString;
}

private static String convertInputStreamToString(InputStream inputStream) throws IOException{
    BufferedReader bufferedReader = new BufferedReader( new InputStreamReader(inputStream));
    String line = "";
    String result = "";
    while((line = bufferedReader.readLine()) != null)
        result += line;

    inputStream.close();
    return result;

}

你可以试试，效果不错

使用 android 上的 HTTPURL 连接从 URL 获取 JSON 对象

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