Question

这应该非常简单，而且我已经接近搞清楚，但我开始放松自己的理智。只是尝试从api中提取数据

如何从对象（另一个对象内）中提取JSON数据？

这是我听说过JSON的第一天，这是第一个承认我对此知之甚少的人。谢谢你的时间！

这篇文章：JSON parsing in android: No value for x error非常有帮助，让我到了现在我正试图打开“列表”然后进入“0”所以我可以收集天气数据。

一直在使用jsonviewer.stack尝试理解这些东西

Api网址：http://api.openweathermap.org/data/2.5/forecast/city?id=4161254&APPID=e661f5bfc93d47b8ed2689f89678a2c9

我的代码：

公共类MainActivity扩展了AppCompatActivity {

TextView citySpot;
TextView cityTemp;

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

    @Override
    protected String doInBackground(String... urls) {

        String result = "";
        URL url;
        HttpURLConnection urlConnection = null;

        try {

            url = new URL(urls[0]);

            urlConnection = (HttpURLConnection) url.openConnection();

            InputStream in = urlConnection.getInputStream();

            InputStreamReader reader = new InputStreamReader(in);

            int data = reader.read();

            while (data != -1) {

                char current = (char) data;

                result += current;

                data = reader.read();

            }
            return result;

        } catch (Exception e) {
            e.printStackTrace();
        }
        return null;

    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);

        try {

            // this works flawlessly and gathers the city name
            JSONObject container = new JSONObject(result);

            JSONObject cityTest = container.getJSONObject("city");

            citySpot.setText(cityTest.getString("name"));

            // this is my failed attempt to get inside of the "list" folder

            JSONObject listList = new JSONObject(result);

            JSONObject listTest = listList.getJSONObject("list");

            JSONObject listTestOne = listTest.getJSONObject("0");

            JSONObject listTestTwo = listTestOne.getJSONObject("main");

            cityTemp.setText(listTestTwo.getString("temp"));
        }
        catch (JSONException e) {
            e.printStackTrace();
        }
    }
}
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);


    citySpot = (TextView) findViewById(R.id.cityName);
    cityTemp = (TextView) findViewById(R.id.textView3);

    DownloadTask task = new DownloadTask();

        task.execute("http://api.openweathermap.org/data/2.5/forecast/city?id=4161254&APPID=e661f5bfc93d47b8ed2689f89678a2c9");


}

}

Answer 1

您可以尝试使用gson库来解析响应。在build.gradle中添加以下内容

compile group: 'com.google.code.gson', name: 'gson', version: '2.3.1'

创建bean以匹配您的响应。使用字段来匹配json响应中返回的内容。以下示例。

public class ResultsResponse {
    private List<MyList> list;
}
public class MyList {
    private String main;
}

如果您希望MyList可以有另一个列表。最后试试

GsonBuilder gsonBuilder =  new GsonBuilder();
ResultsResponse response = gsonBuilder.create().fromJson(jsonString, ResultsResponse.class)

response对象应填充您的列表。

android中的JSON解析：打开对象

1 个答案: