blocks|key|1441187|text|如果你想自己计算，那么你可以使用Haversine公式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1441188|var+rad+=+function(x)+{
++return+x+*+Math.PI+/+180;
};

var+getDistance+=+function(p1,+p2)+{
++var+R+=+6378137;+//+Earth’s+mean+radius+in+meter
++var+dLat+=+rad(p2.lat()+-+p1.lat());
++var+dLong+=+rad(p2.lng()+-+p1.lng());
++var+a+=+Math.sin(dLat+/+2)+*+Math.sin(dLat+/+2)+%2B
++++Math.cos(rad(p1.lat()))+*+Math.cos(rad(p2.lat()))+*
++++Math.sin(dLong+/+2)+*+Math.sin(dLong+/+2);
++var+c+=+2+*+Math.atan2(Math.sqrt(a),+Math.sqrt(1+-+a));
++var+d+=+R+*+c;
++return+d;+//+returns+the+distance+in+meter
};|code-block|syntax|javascript|1441189|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to calculate it yourself, then you can use the Haversine formula:

<pre><code>var rad = function(x) {
 return x * Math.PI / 180;
};

var getDistance = function(p1, p2) {
 var R = 6378137; // Earth’s mean radius in meter
 var dLat = rad(p2.lat() - p1.lat());
 var dLong = rad(p2.lng() - p1.lng());
 var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
 Math.cos(rad(p1.lat())) * Math.cos(rad(p2.lat())) *
 Math.sin(dLong / 2) * Math.sin(dLong / 2);
 var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
 var d = R * c;
 return d; // returns the distance in meter
};
</code></pre>

blocks|key|1441249|text|实际上，在GMap3中似乎有一种方法。它是google.maps.geometry.spherical名称空间的静态方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1441250|它将两个LatLng对象作为参数，并将使用默认地球半径6378137米，但如果需要，可以使用自定义值覆盖默认半径。|1441251|请确保您包括：|1441252|<script+type="text/javascript"+src="http://maps.google.com/maps/api/js?sensor=false&v=3&libraries=geometry"></script>|code-block|syntax|javascript|1441253|在你的头部。|1441254|调用将是：|1441255|google.maps.geometry.spherical.computeDistanceBetween+(latLngA,+latLngB);|1441256|entityMap|0|LINK|mutability|MUTABLE|url|https://developers.google.com/maps/documentation/javascript/reference#spherical^0|L|U|L|U|0|0|4|6|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1B|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1C|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1D|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1E|8|@]|D|@]|E|$]]|$1|S|3|T|5|L|7|1F|8|@]|D|@]|E|$M|N]]|$1|U|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

There actually seems to be a method in GMap3. It's a static method of the <a href="https://developers.google.com/maps/documentation/javascript/reference#spherical"><code>google.maps.geometry.spherical</code></a> namespace. 

It takes as arguments two <code>LatLng</code> objects and will utilize a default Earth radius of 6378137 meters, although the default radius can be overridden with a custom value if necessary.

Make sure you include:

<pre class="lang-html prettyprint-override"><code>&lt;script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&amp;v=3&amp;libraries=geometry"&gt;&lt;/script&gt;
</code></pre>

in your head section.

The call will be:

<pre class="lang-js prettyprint-override"><code>google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);
</code></pre>

blocks|key|1226695|text|使用2点的GPS纬度/经度的示例。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1226696|var+latitude1+=+39.46;
var+longitude1+=+-0.36;
var+latitude2+=+40.40;
var+longitude2+=+-3.68;

var+distance+=+google.maps.geometry.spherical.computeDistanceBetween(new+google.maps.LatLng(latitude1,+longitude1),+new+google.maps.LatLng(latitude2,+longitude2));+++++++|code-block|syntax|javascript|1226697|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Example using GPS latitude/longitude of 2 points.

<pre><code>var latitude1 = 39.46;
var longitude1 = -0.36;
var latitude2 = 40.40;
var longitude2 = -3.68;

var distance = google.maps.geometry.spherical.computeDistanceBetween(new google.maps.LatLng(latitude1, longitude1), new google.maps.LatLng(latitude2, longitude2)); 
</code></pre>

blocks|key|1441357|text|只需将以下代码添加到您的JavaScript代码的开头：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1441358|google.maps.LatLng.prototype.distanceFrom+=+function(latlng)+{
++var+lat+=+[this.lat(),+latlng.lat()]
++var+lng+=+[this.lng(),+latlng.lng()]
++var+R+=+6378137;
++var+dLat+=+(lat[1]-lat[0])+*+Math.PI+/+180;
++var+dLng+=+(lng[1]-lng[0])+*+Math.PI+/+180;
++var+a+=+Math.sin(dLat/2)+*+Math.sin(dLat/2)+%2B
++Math.cos(lat[0]+*+Math.PI+/+180+)+*+Math.cos(lat[1]+*+Math.PI+/+180+)+*
++Math.sin(dLng/2)+*+Math.sin(dLng/2);
++var+c+=+2+*+Math.atan2(Math.sqrt(a),+Math.sqrt(1-a));
++var+d+=+R+*+c;
++return+Math.round(d);
}|code-block|syntax|javascript|1441359|然后像这样使用函数：|1441360|var+loc1+=+new+GLatLng(52.5773139,+1.3712427);
var+loc2+=+new+GLatLng(52.4788314,+1.7577444);
var+dist+=+loc2.distanceFrom(loc1);
alert(dist/1000);|1441361|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Just add this to the beginning of your JavaScript code:

<pre class="lang-js prettyprint-override"><code>google.maps.LatLng.prototype.distanceFrom = function(latlng) {
 var lat = [this.lat(), latlng.lat()]
 var lng = [this.lng(), latlng.lng()]
 var R = 6378137;
 var dLat = (lat[1]-lat[0]) * Math.PI / 180;
 var dLng = (lng[1]-lng[0]) * Math.PI / 180;
 var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
 Math.cos(lat[0] * Math.PI / 180 ) * Math.cos(lat[1] * Math.PI / 180 ) *
 Math.sin(dLng/2) * Math.sin(dLng/2);
 var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
 var d = R * c;
 return Math.round(d);
}
</code></pre>

and then use the function like this:

<pre class="lang-js prettyprint-override"><code>var loc1 = new GLatLng(52.5773139, 1.3712427);
var loc2 = new GLatLng(52.4788314, 1.7577444);
var dist = loc2.distanceFrom(loc1);
alert(dist/1000);
</code></pre>

blocks|key|1441577|text|//p1+and+p2+are+google.maps.LatLng(x,y)+objects

function+calcDistance(p1,+p2)+{
++++++++++var+d+=+(google.maps.geometry.spherical.computeDistanceBetween(p1,+p2)+/+1000).toFixed(2);
++++++++++console.log(d);++++++++++++++
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1441578|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>//p1 and p2 are google.maps.LatLng(x,y) objects

function calcDistance(p1, p2) {
 var d = (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
 console.log(d); 
}
</code></pre>

blocks|key|1226478|text|下面是这个论坛的c#实现|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1226479|+public+class+DistanceAlgorithm
{
++++const+double+PIx+=+3.141592653589793;
++++const+double+RADIO+=+6378.16;

++++///+<summary>
++++///+This+class+cannot+be+instantiated.
++++///+</summary>
++++private+DistanceAlgorithm()+{+}

++++///+<summary>
++++///+Convert+degrees+to+Radians
++++///+</summary>
++++///+<param+name="x">Degrees</param>
++++///+<returns>The+equivalent+in+radians</returns>
++++public+static+double+Radians(double+x)
++++{
++++++++return+x+*+PIx+/+180;
++++}

++++///+<summary>
++++///+Calculate+the+distance+between+two+places.
++++///+</summary>
++++///+<param+name="lon1"></param>
++++///+<param+name="lat1"></param>
++++///+<param+name="lon2"></param>
++++///+<param+name="lat2"></param>
++++///+<returns></returns>
++++public+static+double+DistanceBetweenPlaces(
++++++++double+lon1,
++++++++double+lat1,
++++++++double+lon2,
++++++++double+lat2)
++++{
++++++++double+dlon+=++Radians(lon2+-+lon1);
++++++++double+dlat+=++Radians(lat2+-+lat1);

++++++++double+a+=+(Math.Sin(dlat+/+2)+*+Math.Sin(dlat+/+2))+%2B+Math.Cos(Radians(lat1))+*+Math.Cos(Radians(lat2))+*+(Math.Sin(dlon+/+2)+*+Math.Sin(dlon+/+2));
++++++++double+angle+=+2+*+Math.Atan2(Math.Sqrt(a),+Math.Sqrt(1+-+a));
++++++++return+(angle+*+RADIO)+*+0.62137;//distance+in+miles
++++}

}++++|code-block|syntax|javascript|1226480|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is the c# implementation of the this forumula

<pre><code> public class DistanceAlgorithm
{
 const double PIx = 3.141592653589793;
 const double RADIO = 6378.16;

 /// &lt;summary&gt;
 /// This class cannot be instantiated.
 /// &lt;/summary&gt;
 private DistanceAlgorithm() { }

 /// &lt;summary&gt;
 /// Convert degrees to Radians
 /// &lt;/summary&gt;
 /// &lt;param name="x"&gt;Degrees&lt;/param&gt;
 /// &lt;returns&gt;The equivalent in radians&lt;/returns&gt;
 public static double Radians(double x)
 {
 return x * PIx / 180;
 }

 /// &lt;summary&gt;
 /// Calculate the distance between two places.
 /// &lt;/summary&gt;
 /// &lt;param name="lon1"&gt;&lt;/param&gt;
 /// &lt;param name="lat1"&gt;&lt;/param&gt;
 /// &lt;param name="lon2"&gt;&lt;/param&gt;
 /// &lt;param name="lat2"&gt;&lt;/param&gt;
 /// &lt;returns&gt;&lt;/returns&gt;
 public static double DistanceBetweenPlaces(
 double lon1,
 double lat1,
 double lon2,
 double lat2)
 {
 double dlon = Radians(lon2 - lon1);
 double dlat = Radians(lat2 - lat1);

 double a = (Math.Sin(dlat / 2) * Math.Sin(dlat / 2)) + Math.Cos(Radians(lat1)) * Math.Cos(Radians(lat2)) * (Math.Sin(dlon / 2) * Math.Sin(dlon / 2));
 double angle = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
 return (angle * RADIO) * 0.62137;//distance in miles
 }

} 
</code></pre>

blocks|key|1441434|text|在谷歌，你可以使用spherical+api，google.maps.geometry.spherical.computeDistanceBetween+(latLngA,+latLngB);来做这件事。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1441435|但是，如果球面投影或半正弦解决方案的精度对您来说不够精确(例如，如果您靠近极点或计算更长的距离)，您应该使用不同的库。|1441436|我在维基百科here上找到的关于这个主题的大部分信息。|1441437|要查看任何给定算法的精度是否足够，有一个技巧是填写地球的最大半径和最小半径，并查看差异是否会给您的用例带来问题。有关更多详细信息，请访问this+article|1441438|最后，google+api或haversine将毫无问题地满足大多数用途。|1441439|entityMap|0|LINK|mutability|MUTABLE|url|https://developers.google.com/maps/documentation/javascript/reference?hl=en-US#spherical|1|http://en.wikipedia.org/wiki/Geographic_coordinate_system|2|http://www.cs.nyu.edu/visual/home/proj/tiger/gisfaq.html^0|N|21|9|D|0|0|0|6|4|1|0|1W|C|2|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|16|8|@]|D|@$9|17|A|18|1|19]]|E|$]]|$1|J|3|K|5|6|7|1A|8|@]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|L|3|M|5|6|7|1E|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]|V|$5|Q|R|S|E|$T|W]]|X|$5|Q|R|S|E|$T|Y]]]]

With google you can do it using the <a href="https://developers.google.com/maps/documentation/javascript/reference?hl=en-US#spherical">spherical api</a>, <code>google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);</code>. 

However, if the precision of a spherical projection or a haversine solution is not precise enough for you (e.g. if you're close to the pole or computing longer distances), you should use a different library.

Most information on the subject I found on Wikipedia <a href="http://en.wikipedia.org/wiki/Geographic_coordinate_system">here</a>.

A trick to see if the precision of any given algorithm is adequate is to fill in the maximum and minimum radius of the earth and see if the difference might cause problems for your use case. Many more details can be found in <a href="http://www.cs.nyu.edu/visual/home/proj/tiger/gisfaq.html">this article</a>

In the end the google api or haversine will serve most purposes without problems.

blocks|key|1226541|text|不得不这么做。动作脚本方式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1226542|//just+make+sure+you+pass+a+number+to+the+function+because+it+would+accept+you+mother+in+law...
public+var+rad+=+function(x:*)+{return+x*Math.PI/180;}

protected++function+distHaversine(p1:Object,+p2:Object):Number+{
++++var+R:int+=+6371;+//+earth's+mean+radius+in+km
++++var+dLat:Number+=+rad(p2.lat()+-+p1.lat());
++++var+dLong:Number+=+rad(p2.lng()+-+p1.lng());

++++var+a:Number+=+Math.sin(dLat/2)+*+Math.sin(dLat/2)+%2B
++++++++++++++++Math.cos(rad(p1.lat()))+*+Math.cos(rad(p2.lat()))+*+Math.sin(dLong/2)+*+Math.sin(dLong/2);
++++var+c:Number+=+2+*+Math.atan2(Math.sqrt(a),+Math.sqrt(1-a));
++++var+d:Number+=+R+*+c;

++++return+d;
}|code-block|syntax|javascript|1226543|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Had to do it... The action script way

<pre><code>//just make sure you pass a number to the function because it would accept you mother in law...
public var rad = function(x:*) {return x*Math.PI/180;}

protected function distHaversine(p1:Object, p2:Object):Number {
 var R:int = 6371; // earth's mean radius in km
 var dLat:Number = rad(p2.lat() - p1.lat());
 var dLong:Number = rad(p2.lng() - p1.lng());

 var a:Number = Math.sin(dLat/2) * Math.sin(dLat/2) +
 Math.cos(rad(p1.lat())) * Math.cos(rad(p2.lat())) * Math.sin(dLong/2) * Math.sin(dLong/2);
 var c:Number = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
 var d:Number = R * c;

 return d;
}
</code></pre>

blocks|key|1226669|text|在我的例子中，最好是在SQL+Server中计算这个邮政编码，因为我想获取当前位置，然后搜索距离当前位置一定距离内的所有邮政编码。我也有一个数据库，其中包含邮政编码的列表和他们的最新长度。干杯|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1226670|--will+return+the+radius+for+a+given+number
create+function+getRad(@variable+float)--function+to+return+rad
returns+float
as
begin
declare+@retval+float+
select+@retval=(@variable+*+PI()/180)
--print+@retval
return+@retval
end
go

--calc+distance
--drop+function+dbo.getDistance
create+function+getDistance(@cLat+float,@cLong+float,+@tLat+float,+@tLong+float)
returns+float
as
begin
declare+@emr+float
declare+@dLat+float
declare+@dLong+float
declare+@a+float
declare+@distance+float
declare+@c+float

set+@emr+=+6371--earth+mean+
set+@dLat+=+dbo.getRad(@tLat+-+@cLat);
set+@dLong+=+dbo.getRad(@tLong+-+@cLong);
set+@a+=+sin(@dLat/2)*sin(@dLat/2)%2Bcos(dbo.getRad(@cLat))*cos(dbo.getRad(@tLat))*sin(@dLong/2)*sin(@dLong/2);
set+@c+=+2*atn2(sqrt(@a),sqrt(1-@a))
set+@distance+=+@emr*@c;
set+@distance+=+@distance+*+0.621371+--+i+needed+it+in+miles
--print+@distance
return+@distance;
end+
go


--get+all+zipcodes+within+2+miles,+the+hardcoded+#'s+would+be+passed+in+by+C#
select+*
from+cityzips+a+where+dbo.getDistance(29.76,-95.38,a.lat,a.long)+<3
order+by+zipcode|code-block|syntax|javascript|1226671|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In my case it was best to calculate this in SQL Server, since i wanted to take current location and then search for all zip codes within a certain distance from current location. I also had a DB which contained a list of zip codes and their lat longs. Cheers

<pre><code>--will return the radius for a given number
create function getRad(@variable float)--function to return rad
returns float
as
begin
declare @retval float 
select @retval=(@variable * PI()/180)
--print @retval
return @retval
end
go

--calc distance
--drop function dbo.getDistance
create function getDistance(@cLat float,@cLong float, @tLat float, @tLong float)
returns float
as
begin
declare @emr float
declare @dLat float
declare @dLong float
declare @a float
declare @distance float
declare @c float

set @emr = 6371--earth mean 
set @dLat = dbo.getRad(@tLat - @cLat);
set @dLong = dbo.getRad(@tLong - @cLong);
set @a = sin(@dLat/2)*sin(@dLat/2)+cos(dbo.getRad(@cLat))*cos(dbo.getRad(@tLat))*sin(@dLong/2)*sin(@dLong/2);
set @c = 2*atn2(sqrt(@a),sqrt(1-@a))
set @distance = @emr*@c;
set @distance = @distance * 0.621371 -- i needed it in miles
--print @distance
return @distance;
end 
go


--get all zipcodes within 2 miles, the hardcoded #'s would be passed in by C#
select *
from cityzips a where dbo.getDistance(29.76,-95.38,a.lat,a.long) &lt;3
order by zipcode
</code></pre>

blocks|key|1226675|text|//JAVA
++++public+Double+getDistanceBetweenTwoPoints(Double+latitude1,+Double+longitude1,+Double+latitude2,+Double+longitude2)+{
++++final+int+RADIUS_EARTH+=+6371;

++++double+dLat+=+getRad(latitude2+-+latitude1);
++++double+dLong+=+getRad(longitude2+-+longitude1);

++++double+a+=+Math.sin(dLat+/+2)+*+Math.sin(dLat+/+2)+%2B+Math.cos(getRad(latitude1))+*+Math.cos(getRad(latitude2))+*+Math.sin(dLong+/+2)+*+Math.sin(dLong+/+2);
++++double+c+=+2+*+Math.atan2(Math.sqrt(a),+Math.sqrt(1+-+a));
++++return+(RADIUS_EARTH+*+c)+*+1000;
++++}

++++private+Double+getRad(Double+x)+{
++++return+x+*+Math.PI+/+180;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1226676|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>//JAVA
 public Double getDistanceBetweenTwoPoints(Double latitude1, Double longitude1, Double latitude2, Double longitude2) {
 final int RADIUS_EARTH = 6371;

 double dLat = getRad(latitude2 - latitude1);
 double dLong = getRad(longitude2 - longitude1);

 double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(getRad(latitude1)) * Math.cos(getRad(latitude2)) * Math.sin(dLong / 2) * Math.sin(dLong / 2);
 double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
 return (RADIUS_EARTH * c) * 1000;
 }

 private Double getRad(Double x) {
 return x * Math.PI / 180;
 }
</code></pre>

blocks|key|1441729|text|++/**
+++*+Calculates+the+haversine+distance+between+point+A,+and+B.
+++*+@param+{number[]}+latlngA+[lat,+lng]+point+A
+++*+@param+{number[]}+latlngB+[lat,+lng]+point+B
+++*+@param+{boolean}+isMiles+If+we+are+using+miles,+else+km.
+++*/
++function+haversineDistance(latlngA,+latlngB,+isMiles)+{
++++const+squared+=+x+=>+x+*+x;
++++const+toRad+=+x+=>+(x+*+Math.PI)+/+180;
++++const+R+=+6371;+//+Earth’s+mean+radius+in+km

++++const+dLat+=+toRad(latlngB[0]+-+latlngA[0]);
++++const+dLon+=+toRad(latlngB[1]+-+latlngA[1]);

++++const+dLatSin+=+squared(Math.sin(dLat+/+2));
++++const+dLonSin+=+squared(Math.sin(dLon+/+2));

++++const+a+=+dLatSin+%2B
++++++++++++++(Math.cos(toRad(latlngA[0]))+*+Math.cos(toRad(latlngB[0]))+*+dLonSin);
++++const+c+=+2+*+Math.atan2(Math.sqrt(a),+Math.sqrt(1+-+a));
++++let+distance+=+R+*+c;

++++if+(isMiles)+distance+/=+1.609344;

++++return+distance;
++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1441730|我在网上找到了一个80%25正确的版本，但是插入了错误的参数，并且在使用输入时不一致，这个版本完全解决了这个问题|unstyled|1441731|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code> /**
 * Calculates the haversine distance between point A, and B.
 * @param {number[]} latlngA [lat, lng] point A
 * @param {number[]} latlngB [lat, lng] point B
 * @param {boolean} isMiles If we are using miles, else km.
 */
 function haversineDistance(latlngA, latlngB, isMiles) {
 const squared = x =&gt; x * x;
 const toRad = x =&gt; (x * Math.PI) / 180;
 const R = 6371; // Earth’s mean radius in km

 const dLat = toRad(latlngB[0] - latlngA[0]);
 const dLon = toRad(latlngB[1] - latlngA[1]);

 const dLatSin = squared(Math.sin(dLat / 2));
 const dLonSin = squared(Math.sin(dLon / 2));

 const a = dLatSin +
 (Math.cos(toRad(latlngA[0])) * Math.cos(toRad(latlngB[0])) * dLonSin);
 const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
 let distance = R * c;

 if (isMiles) distance /= 1.609344;

 return distance;
 }
</code></pre>

I found a version online which is 80% right but plugged in the wrong parameter and is inconsistent in using the inputs, this version fixed that completely

blocks|key|1441674|text|要在Google地图上计算距离，可以使用Directions+API。这将是最简单的方法之一。要从Google+Server获取数据，您可以使用Retrofit或Volley。两者都有自己的优势。请看下面的代码，其中我使用了改进来实现它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1441675|private+void+build_retrofit_and_get_response(String+type)+{

++++String+url+=+"https://maps.googleapis.com/maps/";

++++Retrofit+retrofit+=+new+Retrofit.Builder()
++++++++++++.baseUrl(url)
++++++++++++.addConverterFactory(GsonConverterFactory.create())
++++++++++++.build();

++++RetrofitMaps+service+=+retrofit.create(RetrofitMaps.class);

++++Call<Example>+call+=+service.getDistanceDuration("metric",+origin.latitude+%2B+","+%2B+origin.longitude,dest.latitude+%2B+","+%2B+dest.longitude,+type);

++++call.enqueue(new+Callback<Example>()+{
++++++++@Override
++++++++public+void+onResponse(Response<Example>+response,+Retrofit+retrofit)+{

++++++++++++try+{
++++++++++++++++//Remove+previous+line+from+map
++++++++++++++++if+(line+!=+null)+{
++++++++++++++++++++line.remove();
++++++++++++++++}
++++++++++++++++//+This+loop+will+go+through+all+the+results+and+add+marker+on+each+location.
++++++++++++++++for+(int+i+=+0;+i+<+response.body().getRoutes().size();+i%2B%2B)+{
++++++++++++++++++++String+distance+=+response.body().getRoutes().get(i).getLegs().get(i).getDistance().getText();
++++++++++++++++++++String+time+=+response.body().getRoutes().get(i).getLegs().get(i).getDuration().getText();
++++++++++++++++++++ShowDistanceDuration.setText("Distance:"+%2B+distance+%2B+",+Duration:"+%2B+time);
++++++++++++++++++++String+encodedString+=+response.body().getRoutes().get(0).getOverviewPolyline().getPoints();
++++++++++++++++++++List<LatLng>+list+=+decodePoly(encodedString);
++++++++++++++++++++line+=+mMap.addPolyline(new+PolylineOptions()
++++++++++++++++++++++++++++++++++++.addAll(list)
++++++++++++++++++++++++++++++++++++.width(20)
++++++++++++++++++++++++++++++++++++.color(Color.RED)
++++++++++++++++++++++++++++++++++++.geodesic(true)
++++++++++++++++++++);
++++++++++++++++}
++++++++++++}+catch+(Exception+e)+{
++++++++++++++++Log.d("onResponse",+"There+is+an+error");
++++++++++++++++e.printStackTrace();
++++++++++++}
++++++++}

++++++++@Override
++++++++public+void+onFailure(Throwable+t)+{
++++++++++++Log.d("onFailure",+t.toString());
++++++++}
++++});

}|code-block|syntax|javascript|1441676|上面是计算距离的函数build_retrofit_and_get_response的代码。下面是对应的Retrofit接口：|1441677|package+com.androidtutorialpoint.googlemapsdistancecalculator;


import+com.androidtutorialpoint.googlemapsdistancecalculator.POJO.Example;

import+retrofit.Call;
import+retrofit.http.GET;
import+retrofit.http.Query;

public+interface+RetrofitMaps+{


/*
+*+Retrofit+get+annotation+with+our+URL
+*+And+our+method+that+will+return+us+details+of+student.
+*/
@GET("api/directions/json?key=AIzaSyC22GfkHu9FdgT9SwdCWMwKX1a4aohGifM")
Call<Example>+getDistanceDuration(@Query("units")+String+units,+@Query("origin")+String+origin,+@Query("destination")+String+destination,+@Query("mode")+String+mode);|1441678|}|1441679|我希望这能解释你的疑问。万事如意:)|1441680|来源：Google+Maps+Distance+Calculator|offset|length|1441681|entityMap|0|LINK|mutability|MUTABLE|url|http://www.androidtutorialpoint.com/intermediate/google-maps-distance-calculator-using-google-directions-google-maps-android-api/^0|0|0|0|0|0|0|3|V|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@$Q|17|R|18|1|19]]|A|$]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

To calculate distance on Google Maps, you can use Directions API. That will be one of the easiest way to do it. To get data from Google Server, you can use Retrofit or Volley. Both has their own advantage. Take a look at following code where I have used retrofit to implement it:

<pre><code>private void build_retrofit_and_get_response(String type) {

 String url = "https://maps.googleapis.com/maps/";

 Retrofit retrofit = new Retrofit.Builder()
 .baseUrl(url)
 .addConverterFactory(GsonConverterFactory.create())
 .build();

 RetrofitMaps service = retrofit.create(RetrofitMaps.class);

 Call&lt;Example&gt; call = service.getDistanceDuration("metric", origin.latitude + "," + origin.longitude,dest.latitude + "," + dest.longitude, type);

 call.enqueue(new Callback&lt;Example&gt;() {
 @Override
 public void onResponse(Response&lt;Example&gt; response, Retrofit retrofit) {

 try {
 //Remove previous line from map
 if (line != null) {
 line.remove();
 }
 // This loop will go through all the results and add marker on each location.
 for (int i = 0; i &lt; response.body().getRoutes().size(); i++) {
 String distance = response.body().getRoutes().get(i).getLegs().get(i).getDistance().getText();
 String time = response.body().getRoutes().get(i).getLegs().get(i).getDuration().getText();
 ShowDistanceDuration.setText("Distance:" + distance + ", Duration:" + time);
 String encodedString = response.body().getRoutes().get(0).getOverviewPolyline().getPoints();
 List&lt;LatLng&gt; list = decodePoly(encodedString);
 line = mMap.addPolyline(new PolylineOptions()
 .addAll(list)
 .width(20)
 .color(Color.RED)
 .geodesic(true)
 );
 }
 } catch (Exception e) {
 Log.d("onResponse", "There is an error");
 e.printStackTrace();
 }
 }

 @Override
 public void onFailure(Throwable t) {
 Log.d("onFailure", t.toString());
 }
 });

}
</code></pre>

Above is the code of function build_retrofit_and_get_response for calculating distance. Below is corresponding Retrofit Interface:

<pre><code>package com.androidtutorialpoint.googlemapsdistancecalculator;


import com.androidtutorialpoint.googlemapsdistancecalculator.POJO.Example;

import retrofit.Call;
import retrofit.http.GET;
import retrofit.http.Query;

public interface RetrofitMaps {


/*
 * Retrofit get annotation with our URL
 * And our method that will return us details of student.
 */
@GET("api/directions/json?key=AIzaSyC22GfkHu9FdgT9SwdCWMwKX1a4aohGifM")
Call&lt;Example&gt; getDistanceDuration(@Query("units") String units, @Query("origin") String origin, @Query("destination") String destination, @Query("mode") String mode);
</code></pre>

}

I hope this explains your query. All the best :)

Source: <a href="http://www.androidtutorialpoint.com/intermediate/google-maps-distance-calculator-using-google-directions-google-maps-android-api/" rel="nofollow">Google Maps Distance Calculator</a>

blocks|key|1441769|text|首先，你是指整个路径的长度的距离，还是只想知道位移(直线距离)？我看这里没有人指出距离和位移之间的区别。对于由JSON/XML数据给出的每个路由点的距离计算，对于位移，有一个使用Spherical类的内置解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1441770|//calculates+distance+between+two+points+in+km's
function+calcDistance(p1,+p2)+{
++return+(google.maps.geometry.spherical.computeDistanceBetween(p1,+p2)+/+1000).toFixed(2);
}|code-block|syntax|javascript|1441771|entityMap^0|2H|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

First, are you referring to distance as in length of the entire path or you want to know only the displacement (straight line distance)? I see no one is pointing the difference between distance and displacement here. For distance calculate each route point given by JSON/XML data, as for displacement there is a built-in solution using <code>Spherical</code> class
<pre><code>//calculates distance between two points in km's
function calcDistance(p1, p2) {
 return (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
}
</code></pre>

How do you calculate the distance between two markers in Google maps V3? (Similar to the <code>distanceFrom</code> function inV2.)

Thanks..

Calculate distance between two points in google maps V3

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如何计算谷歌地图V3中两个标记之间的距离？(类似于distanceFrom函数inV2。)谢谢..

问在谷歌地图V3中计算两点之间的距离
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回答 13

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