使用Google地图在定义区域列出事件 [英] List Events on defined area using Google Maps

问题描述

我有谷歌地图定义了很多事件等。

我如何列出所有事件，例如。因为你的问题有点含糊，我假设你有一个lat，long坐标列表。您可以使用类似下面的方法将它们从SQL中取出：

  SELECT 
 z.id AS z__id，
 z.zipcode AS z__zipcode，
 z.city AS z__city，
 z.state AS z_state，
 z.county AS z__county，
 z.zip_class AS z__zip_class，
（纬度/海平面/ 180）* SIN（z.latitude * PI（）/ 180）+ COS（纬度/海平面） *（180度）* 180度（180度）* 180度* 180度）* COS（z.latitude * PI（）/ 180）* COS（（ -  z.longitude）* PI（）/ 180） （）（）（）（）（）（）（）（）（）（）（） （）/ 180）* COS（（ -  z.longitude）* PI（）/ 180））* 180 / PI（））* 60 * 1.1515 * 1.609344）AS z__1 
 FROM zipcode z 
 WHERE z.city！=？ 
 ORDER BY z__0 asc 
 LIMIT 50 
  

z_ 0将会（以千米为单位），而z-1将以千米为单位的距离。 b
$ b

（来源： http://www.doctrine-project.org/projects/orm/1.2/docs/手动/行为/ EN＃核心行为：地理上的）

或者使用类似的东西（类似于php）：

 函数距离（$ lat1，$ lon1，$ lat2，$ lon2，$ unit）{
 
 $ theta = $ lon1  -  $ lon2; 
 $ dist = sin（deg2rad（$ lat1））* sin（deg2rad（$ lat2））+ cos（deg2rad（$ lat1））* cos（deg2rad（$ lat2））* cos（deg2rad（$ theta） ）; 
 $ dist = acos（$ dist）; 
 $ dist = rad2deg（$ dist）; 
 $ miles = $ dist * 60 * 1.1515; 
 $ unit = strtoupper（$ unit）; 
 
 if（$ unit ==K）{
 return（$ miles * 1.609344）; 
} else if（$ unit ==N）{
 return（$ miles * 0.8684）; 
} else {
 return $ miles; 
} 
} 
  

http://www.zipcodeworld.com/samples/distance.php.html

I have Google Map with defined a lot of events etc.

How I can list all events eg. in 50km around my location?

解决方案

Since your question is a bit vague, I assume you have a list of lat, long coordinates. You can get them out of SQL using something like this:

SELECT 
z.id AS z__id, 
z.zipcode AS z__zipcode, 
z.city AS z__city, 
z.state AS z__state, 
z.county AS z__county, 
z.zip_class AS z__zip_class, 
z.latitude AS z__latitude, 
z.longitude AS z__longitude, 
((ACOS(SIN(* PI() / 180) * SIN(z.latitude * PI() / 180) + COS(* PI() / 180) * COS(z.latitude * PI() / 180) * COS((- z.longitude) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS z__0, 
((ACOS(SIN(* PI() / 180) * SIN(z.latitude * PI() / 180) + COS(* PI() / 180) * COS(z.latitude * PI() / 180) * COS((- z.longitude) * PI() / 180)) * 180 / PI()) * 60 * 1.1515 * 1.609344) AS z__1 
FROM zipcode z 
WHERE z.city != ? 
ORDER BY z__0 asc 
LIMIT 50

z_0 will be the dinstance in miles, whereas z_1 will be the distance in km.

(source: http://www.doctrine-project.org/projects/orm/1.2/docs/manual/behaviors/en#core-behaviors:geographical)

Or use something like this (similar in php):

function distance($lat1, $lon1, $lat2, $lon2, $unit) { 

  $theta = $lon1 - $lon2; 
  $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta)); 
  $dist = acos($dist); 
  $dist = rad2deg($dist); 
  $miles = $dist * 60 * 1.1515;
  $unit = strtoupper($unit);

  if ($unit == "K") {
    return ($miles * 1.609344); 
  } else if ($unit == "N") {
      return ($miles * 0.8684);
    } else {
        return $miles;
      }
}

http://www.zipcodeworld.com/samples/distance.php.html

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