将 Json 数据解析为 Json 对象答案

【问题标题】：Parse Json data into a Json object将 Json 数据解析为 Json 对象
【发布时间】：2019-02-08 12:11:13
【问题描述】：

我在解析 Json 对象内的标签时遇到问题。我的 json 代码的结构是这样的：

{"giocatori":[{"nome":"Giovanni","cognome":"Muchacha","numero":"1","ruolo":"F-G"},
{"nome":"Giorgio","cognome":"Rossi","numero":"2","ruolo":"AG"},
{"nome":"Andrea","cognome":"Suagoloso","numero":"3","ruolo":"P"},
{"nome":"Salvatore","cognome":"Aranzulla","numero":"4","ruolo":"G"},
{"nome":"Giulio","cognome":"Muchacha","numero":"5","ruolo":"F"}]}

我得到了让我从这里获取 Json 文件的代码：Get JSON Data from URL Using Android?，我正在尝试将标签（例如“nome”标签）解析为 Json 对象。这是我得到的代码：

public class MainActivity extends AppCompatActivity {

Button btnHit;
TextView txtJson;
ProgressDialog pd;

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);

btnHit = (Button) findViewById(R.id.btnHit);
txtJson = (TextView) findViewById(R.id.tvJsonItem);

btnHit.setOnClickListener(new View.OnClickListener() {
    @Override
    public void onClick(View v) {
        new JsonTask().execute("https://api.myjson.com/bins/177dpo");
    }
 });


}


private class JsonTask extends AsyncTask<String, String, String> {

protected void onPreExecute() {
    super.onPreExecute();

    pd = new ProgressDialog(MainActivity.this);
    pd.setMessage("Please wait");
    pd.setCancelable(false);
    pd.show();
}

protected String doInBackground(String... params) {


    HttpURLConnection connection = null;
    BufferedReader reader = null;

    try {
        URL url = new URL(params[0]);
        connection = (HttpURLConnection) url.openConnection();
        connection.connect();


        InputStream stream = connection.getInputStream();

        reader = new BufferedReader(new InputStreamReader(stream));

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

        while ((line = reader.readLine()) != null) {
            buffer.append(line+"\n");
            Log.d("Response: ", "> " + line);   

        }

        return buffer.toString();


    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        if (connection != null) {
            connection.disconnect();
        }
        try {
            if (reader != null) {
                reader.close();
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return null;
   }

@Override
protected void onPostExecute(String result) {
    super.onPostExecute(result);
    if (pd.isShowing()){
        pd.dismiss();
    }
    txtJson.setText(result);
 }
}
}

我从未使用过这种类型的文件，因此非常感谢您的帮助！

【问题讨论】：

可以使用 jsonschema2pojo.com 生成 POJO 类
您能否制作一个示例代码以便我理解您的意思并将问题标记为已回答，谢谢
只需复制您的 gson 并将其粘贴到那里，我现在在防火墙后面，无法访问该站点。

标签： android json

【解决方案1】：

你可以这样使用：

try {
                    String servResponse = response.toString();
                    JSONObject parentObj = new JSONObject(servResponse);
                    JSONArray parentArray = parentObj.getJSONArray("giocatori");

                    if (parentArray.length() == 0) {
                        //if it's empty, do something (or not)

                    } else {
                        //Here, finalObj will have your jsonObject
                        JSONObject finalObj = parentArray.getJSONObject(0);
                        //if you decide to store some value of the object, you can do like this (i've created a nomeGiocatori for example)
                        nomeGiocatori = finalObj.getString("nome");

                    }

                } catch (Exception e) {
                    Log.d("Exception: ", "UnknownException");
                }

我一直使用这种代码，就像一个魅力。

【讨论】：