Android / JSON 从同一活动中的两个 JSON URL 返回对象

Question

这是代码：

@Override
        protected DVLAInformation doInBackground(String... params) {
            try {

                Intent intent = getIntent();
                String dvlaNumFin = intent.getStringExtra("dvlaNumber");

                final URL url = new URL("https://dvlasearch.appspot.com/DvlaSearch?licencePlate=" + dvlaNumFin + "&apikey=");
                HttpURLConnection connection = (HttpURLConnection) url.openConnection();


                connection.setRequestMethod("GET");
                connection.setRequestProperty("USER-AGENT", "Mozilla/5.0");
                connection.setRequestProperty("ACCEPT-LANGUAGE", "en-US,en;0.5");

                BufferedReader br = new BufferedReader(new InputStreamReader(connection.getInputStream()));
                String line;
                StringBuilder responseOutput = new StringBuilder();
                while ((line = br.readLine()) != null) {
                    responseOutput.append(line);
                }
                br.close();

                DVLAInformation obj = new DVLAInformation(url.toString());
                obj.readAndParseJSON(responseOutput.toString());



                return obj;


            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            return null;


        }

我的问题是如何在同一活动中从两个 JSON URL 返回对象

例如我会有两个网址

final URL url = new URL("https://dvlasearch.appspot.com/DvlaSearch?licencePlate=" + dvlaNumFin + "&apikey=DvlaSearchDemoAccount");
                HttpURLConnection connection = (HttpURLConnection) url.openConnection();

final URL url2 = new URL("https://dvlasearch.appspot.com/MotHistory?licencePlate=mt09nks&apikey=DvlaSearchDemoAccount");
                HttpURLConnection connection = (HttpURLConnection) url.openConnection();

而且我不知道如何返回这样的东西：

    DVLAInformation obj = new DVLAInformation(url.toString());
                    obj.readAndParseJSON(responseOutput.toString());
    DVLAInformation obj2 = new DVLAInformation(url2.toString());
                    obj2.readAndParseJSON(responseOutput.toString());

return obj, obj2

抱歉我的解释，我知道方法不对，我只是试着解释一下我想做什么。

找到这个CLICK，但不明白如何应用它

Answer 1

看看Retrofit。如果您需要指导，您可能会发现这个Tutorial 很有用。

作为替代方案，您可以使用 OkHttp 并链接 HTTP 请求。