java - 解析 JSON 数组和对象

Question

我想通过 API 检索足球运动员列表，我已正确进行 Http 处理程序和 api 调用。现在我有了这个 JSON 数组

http://api.football-data.org/v1/teams/66/players

我想对其进行解析，以便仅显示玩家的姓名。我如何解析 JSON 数组的第一位，以便数组从 [{name:Paul Pogba... 开始？

到目前为止我的代码:

@Override
    protected Void doInBackground(Void... arg0) {
        //New instance of http
        http sh = new http();

        // Making a request to URL and getting response
        final String jsonStr = sh.makeServiceCall(url);

        Log.e(TAG, "Response from: " + jsonStr);

        if (jsonStr != null) {
            try {
                // Getting JSON Array node
                JSONArray jsonarray = new JSONArray(jsonStr);


                for (int i = 0; i < jsonarray.length(); i++) {
                    JSONObject jo = jsonarray.getJSONObject(i);

                    String name = jo.getString("name");

                    HashMap<String, String> player = new HashMap<>();

                    player.put("name", name);

                    playerlist.add(player);
                }
            } catch (final JSONException e) { //In case an error regarding JSON parsing takes place
                Log.e(TAG, "Json parsing error: " + e.getMessage());
                runOnUiThread(new Runnable() {
                    @Override
                    public void run() {
                        Toast.makeText(getApplicationContext(),
                                "Json parsing error: " + e.getMessage(),
                                Toast.LENGTH_LONG)
                                .show();
                    }
                });

            }
        } else {
            Log.e(TAG, "Couldn't get json from server."); //In case the JSON can't be obtained from the server
            runOnUiThread(new Runnable() {
                @Override
                public void run() {
                    Toast.makeText(getApplicationContext(),
                            "Couldn't get JSON from server. Check LogCat for possible errors!",
                            Toast.LENGTH_LONG)
                            .show();
                }
            });

        }

        return null;
    }

Answer 1

谢谢各位提供的信息。用子字符串修复了它:)