如何从使用经纬度坐标的位置获得时区?

给定一个位置的经度和纬度，如何知道该位置的有效时区?

在大多数情况下，我们正在寻找IANA/Olson时区id，尽管有些服务可能只返回UTC偏移量或其他一些时区标识符。详细信息请阅读时区标签信息。

当前回答

2019-12-02 05:23:19

其他回答

下面是如何使用谷歌的脚本编辑器来获取gsheet中的timezoneName和timeZoneId。

步骤1。获取谷歌的时区API的API键

步骤2。创建一个新的gsheet。在“工具”菜单下单击“脚本编辑器”。添加如下代码:

function getTimezone(lat, long) {  
  var apiKey = 'INSERTAPIKEYHERE'
  var url = 'https://maps.googleapis.com/maps/api/timezone/json?location=' + lat + ',' + long + '&timestamp=1331161200&key=' + apiKey 
  var response = UrlFetchApp.fetch(url);
  var data = JSON.parse(response.getContentText());
  return data["timeZoneName"];
}

步骤3。保存并发布getTimezone()函数，并如上图所示使用它。

2017-12-08 16:34:35

https://en.wikipedia.org/wiki/Great-circle_distance

下面是一个使用JSON数据的很好的实现: https://github.com/agap/llttz

public TimeZone nearestTimeZone(Location node) {
    double bestDistance = Double.MAX_VALUE;
    Location bestGuess = timeZones.get(0);

    for (Location current : timeZones.subList(1, timeZones.size())) {
        double newDistance = distanceInKilometers(node, current);

        if (newDistance < bestDistance) {
            bestDistance = newDistance;
            bestGuess = current;
        }
    }

    return java.util.TimeZone.getTimeZone(bestGuess.getZone());
}

  protected double distanceInKilometers(final double latFrom, final double lonFrom, final double latTo, final double lonTo) {
        final double meridianLength = 111.1;
        return meridianLength * centralAngle(latFrom, lonFrom, latTo, lonTo);
    }

    protected double centralAngle(final Location from, final Location to) {
        return centralAngle(from.getLatitude(), from.getLongitude(), to.getLatitude(), to.getLongitude());
    }

    protected double centralAngle(final double latFrom, final double lonFrom, final double latTo, final double lonTo) {
        final double latFromRad = toRadians(latFrom),
                lonFromRad = toRadians(lonFrom),
                latToRad   = toRadians(latTo),
                lonToRad   = toRadians(lonTo);

        final double centralAngle = toDegrees(acos(sin(latFromRad) * sin(latToRad) + cos(latFromRad) * cos(latToRad) * cos(lonToRad - lonFromRad)));

        return centralAngle <= 180.0 ? centralAngle : (360.0 - centralAngle);
    }

    protected double distanceInKilometers(final Location from, final Location to) {
        return distanceInKilometers(from.getLatitude(), from.getLongitude(), to.getLatitude(), to.getLongitude());
    }
}

2015-08-19 10:44:07

通过使用纬度和经度得到当前位置下面的时区代码为我工作

String data = null;         
LocationManager locationManager = (LocationManager) getSystemService(LOCATION_SERVICE);
Location ll = locationManager.getLastKnownLocation(LocationManager.NETWORK_PROVIDER);
double lat = 0,lng = 0;
if(ll!=null){
    lat=ll.getLatitude();
    lng=ll.getLongitude();
}
System.out.println(" Last known location of device  == "+lat+"    "+lng);

InputStream iStream = null; 
HttpURLConnection urlConnection = null;
try{
    timezoneurl = timezoneurl+"location=22.7260783,75.8781553&timestamp=1331161200";                    
    // timezoneurl = timezoneurl+"location="+lat+","+lng+"&timestamp=1331161200";

    URL url = new URL(timezoneurl);                
    // Creating an http connection to communicate with url 
    urlConnection = (HttpURLConnection) url.openConnection(); 

    // Connecting to url 
    urlConnection.connect();                

    // Reading data from url 
    iStream = urlConnection.getInputStream();

    BufferedReader br = new BufferedReader(new InputStreamReader(iStream));

    StringBuffer sb  = new StringBuffer();
    String line = "";
    while( ( line = br.readLine())  != null){
        sb.append(line);
    }
    data = sb.toString();
    br.close();

}catch(Exception e){
    Log.d("Exception while downloading url", e.toString());
}finally{
    try {
        iStream.close();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    urlConnection.disconnect();
}

try {
    if(data!=null){
        JSONObject jobj=new JSONObject(data);
        timezoneId = jobj.getString("timeZoneId");

        SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd hh:mm:ss");
        format.setTimeZone(TimeZone.getTimeZone(timezoneId));

        Calendar cl = Calendar.getInstance(TimeZone.getTimeZone(timezoneId));
        System.out.println("time zone id in android ==  "+timezoneId);

        System.out.println("time zone of  device in android == "+TimeZone.getTimeZone(timezoneId));
        System.out.println("time fo device in android "+cl.getTime());
    }
} catch (Exception e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

2014-11-06 12:37:24

There are several sources online that have geojson data for timezones (here's one, here's another) Use a geometry library to create polygon objects from the geojson coordinates (shapely [python], GEOS [c++], JTS [java], NTS [.net]). Convert your lat/lng to a point object (however your library represents that) and check if it intersects the timezone polygon. from shapely.geometry import Polygon, Point def get_tz_from_lat_lng(lat, lng): for tz, geojson in timezones.iteritems(): coordinates = geojson['features'][0]['geometry']['coordinates'] polygon = Polygon(coordinates) point = Point(lng, lat) if polygon.contains(point): return tz

2017-01-13 19:10:22

node.js的这个解决方案怎么样 https://github.com/mattbornski/tzwhere

和它的Python对等体: https://github.com/pegler/pytzwhere

2014-03-05 17:21:16

如何从使用经纬度坐标的位置获得时区?

推荐文章

最新文章

标签