【问题标题】：Get weather for specific date using Openweather API - parsing JSON response使用 Openweather API 获取特定日期的天气 - 解析 JSON 响应
【发布时间】：2020-03-16 16:14:14
【问题描述】：

我正在尝试使用 Android Studio 和 Java 构建一个简单的天气预报应用程序。我按照这里的一些说明（https://www.androdocs.com/java/creating-an-android-weather-app-using-java.html）启动并运行，这很有效。但是，我只能得到当前的天气。 Openweather 预测 API 调用似乎持续了 5 天。没关系，但是我如何获取用户指定的未来 5 天内的特定日期的天气（比如温度和风速）？

以下是示例 JSON 响应（已缩短）。即使我可以在下午 12 点提取特定日期的信息，并获得该日期的温度和风速，也足够了。如何解析此 JSON 响应以获取特定日期的温度和风速？非常感谢...对不起我是初学者...

{"cod":"200","message":0,"cnt":40,"list":[{"dt":1574283600,"main":{"temp":281.75,"temp_min" :281.68,"temp_max":281.75,"压力":995,"sea_level":995,"grnd_level":980,"湿度":93,"temp_kf":0.07},"天气":[{"id": 501,"main":"Rain","description":"中雨","icon":"10n"}],"clouds":{"all":100},"wind":{"speed": 4.82,"deg":147},"rain":{"3h":5.38},"sys":{"pod":"n"},"dt_txt":"2019-11-20 21:00:00 "},{"dt":1574294400,"main":{"temp":281.79,"temp_min":281.74,"temp_max":281.79,"pressure":995,"sea_level":995,"grnd_level":980 ,"湿度":91,"temp_kf":0.05},"天气":[{"id":500,"main":"Rain","description":"小雨","icon":"10n" }],"clouds":{"all":100},"wind":{"speed":5.55,"deg":140},"rain":{"3h":1.75},"sys":{ "pod":"n"},"dt_txt":"2019-11-21 00:00:00"},{"dt":1574305200,"main":{"temp":279.48,"temp_min":279.44 ,"temp_max":279.48,"压力":994,"sea_level":994,"grnd_level":980,"湿度":95,"temp_kf":0.04},"天气":[{"id":500, "main":"Rain","description":"小雨","icon":"10n"}],"clouds":{"all":10 0},"wind":{"speed":2.37,"deg":155},"rain":{"3h":0.94},"sys":{"pod":"n"},"dt_txt" :"2019-11-21 03:00:00"},{"dt":1574316000,"main":{"temp":278.56,"temp_min":278.54,"temp_max":278.56,"pressure":995 ,"sea_level":995,"grnd_level":980,"湿度":94,"temp_kf":0.02},"weather":[{"id":500,"main":"Rain","description": "小雨","icon":"10n"}],"clouds":{"all":100},"wind":{"speed":1.73,"deg":128},"rain":{ "3h":0.06},"sys":{"pod":"n"},"dt_txt":"2019-11-21 06:00:00"},{"dt":1574326800,"main": {"temp":279.19,"temp_min":279.19,"temp_max":279.19,"pressure":995,"sea_level":995,"grnd_level":981,"湿度":95,"temp_kf":0}, "天气":[{"id":804,"main":"Clouds","description":"阴云","icon":"04d"}],"clouds":{"all":100} ,"wind":{"speed":1.79,"deg":104},"sys":{"pod":"d"},"dt_txt":"2019-11-21 09:00:00"} ,{"dt":1574337600,"main":{"temp":282.2,"temp_min":282.2,"temp_max":282.2,"pressure":995,"sea_level":995,"grnd_level":980,"湿度":85,"temp_kf":0},"天气":[{"id":500,"main":"Rain","description":"小雨","icon":"10d"}],"clouds":{"all":100},"wind":{"speed":2.78,"deg":129},"rain":{"3h" :0.19},"sys":{"pod":"d"},"dt_txt":"2019-11-21 12:00:00"}

【问题讨论】：

你试过用gson吗？ futurestud.io/tutorials/…
链接到您的数据源。
api.openweathermap.org/data/2.5/forecast?q=Cork&appid=3d53978165d565c8523a9d25b1fa9c60 - 您好，我的数据来源是这个 API 调用。感谢您的 cmets。

标签： java json android-studio parsing weather-api

【解决方案1】：

JSON 简单

这是一个使用 JSON-Simple 库解析从 OpenWeatherMap.org 下载的 JSON 数据的示例应用。

package work.basil.example;

import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;
import org.json.simple.parser.ParseException;

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URL;
import java.net.URLConnection;
import java.time.Instant;
import java.time.ZoneId;
import java.time.ZonedDateTime;

public class Weather
{

    public static void main ( String[] args )
    {

        Weather app = new Weather();
        app.demo();
    }

    private void demo ( )
    {
        //Creating a JSONParser object
        JSONParser jsonParser = new JSONParser();
        try
        {
            // Download JSON.
            String yourKey = "b6907d289e10d714a6e88b30761fae22";
            URL url = new URL( "https://samples.openweathermap.org/data/2.5/forecast/hourly?zip=79843&appid=b6907d289e10d714a6e88b30761fae22" + yourKey ); // 79843 = US postal Zip Code for Marfa, Texas.
            URLConnection conn = url.openConnection();
            BufferedReader reader = new BufferedReader( new InputStreamReader( conn.getInputStream() ) );


            // Parse JSON
            JSONObject jsonObject = ( JSONObject ) jsonParser.parse( reader );
            System.out.println( "jsonObject = " + jsonObject );

            JSONArray list = ( JSONArray ) jsonObject.get( "list" );
            System.out.println( "list = " + list );

            // Loop through each item
            for ( Object o : list )
            {
                JSONObject forecast = ( JSONObject ) o;

                Long dt = ( Long ) forecast.get( "dt" );          // Parse text into a number of whole seconds.
                Instant instant = Instant.ofEpochSecond( dt );    // Parse the count of whole seconds since 1970-01-01T00:00Z into a `Instant` object, representing a moment in UTC with a resolution of nanoseconds.
                ZoneId z = ZoneId.of( "America/Chicago" );        // Specify a time zone using a real `Continent/Region` time zone name. Never use 2-4 letter pseudo-zones such as `PDT`, `CST`, `IST`, etc.
                ZonedDateTime zdt = instant.atZone( z );          // Adjust from a moment in UTC to the wall-clock used by the people of a particular region (a time zone). Same moment, same point on the timeline, different wall-clock time.
                LocalTime lt = zdt.toLocalTime() ;
                // … compare with lt.equals( LocalTime.NOON ) to find the data sample you desire. 
                System.out.println( "dt : " + dt );
                System.out.println( "instant : " + instant );
                System.out.println( "zdt : " + zdt );

                JSONObject main = ( JSONObject ) forecast.get( "main" );
                System.out.println( "main = " + main );


                Double temp = ( Double ) main.get( "temp" );  // Better to use BigDecimal instead of Double for accuracy. But I do not know how to get the JSON-Simple library to parse the original string input as a BigDecimal.
                System.out.println( "temp = " + temp );

                JSONObject wind = ( JSONObject ) forecast.get( "wind" );
                System.out.println( "wind = " + wind );

                System.out.println( "BASIL - wind.getCLass: " + wind.getClass() );
                Double speed = ( Double ) wind.get( "speed" );
                System.out.println( "speed = " + speed );

                System.out.println( "\n" );
            }
        }
        catch ( FileNotFoundException e )
        {
            e.printStackTrace();
        }
        catch ( IOException e )
        {
            e.printStackTrace();
        }
        catch ( ParseException e )
        {
            e.printStackTrace();
        }
    }
}

小数分隔符

请注意，当遇到缺少小数分隔符的风速数据点时，此代码会崩溃。例如，该数据的发布者应写入1.0 而不是1 以保持一致性。如果他们这样做了，库会将1.0 解析为Double，而不是将1 解析为Long。

JSON-Simple 1 现已失效

此外，此代码使用 JSON-Simple 的原始版本 1，现已失效。这个项目是分叉的，产生了截然不同的版本 2 和 3。

有关解析十进制问题的详细信息以及指向forked project 的链接，请参阅此页面Parsing decimal numbers, some of which lack a decimal separator, in JSON data using JSON-Simple (Java)。

不可用于生产

因此，虽然我不推荐将此代码用于生产用途，但它可能会对您有所帮助。对于实际工作，请考虑 JSON-Simple 的更高版本 3 或其他几个可用于 Java 的 JSON 处理库中的任何一个。

参见this URL 的示例数据。要使其可读，请使用您的文本编辑器或 IDE 重新格式化 JSON 数据。

样本输出：

dt : 1553709600
instant : 2019-03-27T18:00:00Z
zdt : 2019-03-27T13:00-05:00[America/Chicago]
main = {"temp":286.44,"temp_min":286.258,"grnd_level":1002.193,"temp_kf":0.18,"humidity":100,"pressure":1015.82,"sea_level":1015.82,"temp_max":286.44}
temp = 286.44
wind = {"deg":202.816,"speed":5.51}
speed = 5.51


dt : 1553713200
instant : 2019-03-27T19:00:00Z
zdt : 2019-03-27T14:00-05:00[America/Chicago]
main = {"temp":286.43,"temp_min":286.3,"grnd_level":1002.667,"temp_kf":0.13,"humidity":100,"pressure":1016.183,"sea_level":1016.183,"temp_max":286.43}
temp = 286.43
wind = {"deg":206.141,"speed":4.84}
speed = 4.84

【讨论】：

【解决方案2】：

您可以通过任何一种最流行的 JSON 库（例如 Jackson 或 Gson）将响应 JSON 字符串序列化为 POJO，然后检索日期字段等于给定日期的对象的所需字段。顺便说一句，您的 JSON 字符串无效，]} 在其末尾丢失。

POJO

@JsonIgnoreProperties(ignoreUnknown = true)
class Response {
    List<Weather> list;

    //general getters and setters
}

@JsonIgnoreProperties(ignoreUnknown = true)
class Weather {
    JsonNode main;
    JsonNode wind;

    @JsonProperty("dt_txt")
    String dtTxt;

    //general getters and setters
}

使用@JsonIgnoreProperties（Jackson 提供）忽略序列化时不关心的那些字段。

代码 sn-p

ObjectMapper mapper = new ObjectMapper();
Response response = mapper.readValue(jsonStr, Response.class);

String givenDate = "2019-11-21 12:00:00";
response.getList().forEach(e -> {
    if (givenDate.equals(e.getDtTxt())) {
        System.out.println("temp: " + e.getMain().get("temp").asText());
        System.out.println("wind speed:" + e.getWind().get("speed").asText());
    }
});

控制台输出

温度：282.2
风速：2.78

【讨论】：