从Retrofit响应解析JSON对象

Question

我是android dev的新手，我首先开发了一个天气应用程序，它可以显示天气预报和当前天气预报。我想添加一个允许用户查看每日天气预报的按钮，但从API调用看，它似乎返回了一个JSON对象。我已经被困在这个问题上好几天了，我不知道该怎么办。

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        Retrofit retrofit = new Retrofit.Builder()
                .baseUrl("https://api.weatherapi.com/v1/")
                .addConverterFactory(GsonConverterFactory.create())
                .build();
        APICall apiCall = retrofit.create(APICall.class);
        Call<WeatherModel> call = apiCall.getHourAndAstroDetails(latlong, "7");
        System.out.println(call.request().url());
        call.enqueue(new Callback<WeatherModel>() {
            @Override
            public void onResponse(Call<WeatherModel> call, Response<WeatherModel> response) {
                if (days == 1) {
                    weatherList = response.body().getForecast().getForecastday().get(0).getHour();
                    System.out.println("this is the size of weatherList" + weatherList.size());
                    setAdapter(weatherList);
                } else if (days == 7) {
                    dailyForcast = response.body().getForecast().getForecastday().get(0).getDay();
                }


                String sunriseTime = response.body().getForecast().getForecastday().get(0).getAstro().getSunrise();
                String sunsetTime = response.body().getForecast().getForecastday().get(0).getAstro().getSunset();
                sunriseTextView.setText(sunriseTime);
                sunsetTextView.setText(sunsetTime);
            }

            @Override
            public void onFailure(Call<WeatherModel> call, Throwable t) {
            }
        });
    }

Answer 1

我认为您应该为从api获得的所有数据创建一个模型类，这样您就可以更容易地处理数据。

这篇文章解释了，

https://medium.com/swlh/basic-usage-of-retrofit-in-kotlin-for-fetching-json-8b9d37999058