blocks|key|4038968|text|您可以使用Google地图获取地址的经度/纬度请参阅Thomas+Clayson的回答|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4038969|How+can+I+get+latitude,+longitude+of+a+location+programmatically+or+using+a+api|offset|length|4038970|然后，您可以使用余弦定律或哈维正弦公式See+Law+of+Cosines计算两组坐标之间的距离|4038971|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/3617227/how-can-i-get-latitude-longitude-of-a-location-programmatically-or-using-a-api|1|http://blog.falafel.com/blogs/josh-eastburn/2012/03/30/finding-closest-location-by-latitude-and-longitude^0|0|0|27|0|0|J|I|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@$D|X|E|Y|1|Z]]|A|$]]|$1|H|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]|P|$5|K|L|M|A|$N|Q]]]]

You can use Google Maps to get Lat/Long of address
See answer by Thomas Clayson

<a href="https://stackoverflow.com/questions/3617227/how-can-i-get-latitude-longitude-of-a-location-programmatically-or-using-a-api">How can I get latitude, longitude of a location programmatically or using a api</a>

Then you can calculate the distance between the two sets of coordinates by using the Law of Cosines or the Haversine formula
<a href="http://blog.falafel.com/blogs/josh-eastburn/2012/03/30/finding-closest-location-by-latitude-and-longitude" rel="nofollow noreferrer">See Law of Cosines</a>

blocks|key|259007|text|这取决于您已经在使用的其他服务。例如，我们已经使用google地图来显示路线，因此我们使用它们的geocoding+service将地址转换为坐标，并使用their+distance+service来计算距离。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|259008|entityMap|0|LINK|mutability|MUTABLE|url|https://developers.google.com/maps/documentation/javascript/examples/geocoding-simple|1|https://developers.google.com/maps/documentation/javascript/examples/distance-matrix^0|1C|H|0|25|M|1|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]|$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]]]

It depends on the other services you are already using. For example, we already use google maps to display the route, so we use their <a href="https://developers.google.com/maps/documentation/javascript/examples/geocoding-simple">geocoding service</a> to translate from an address to coordinates, as well as <a href="https://developers.google.com/maps/documentation/javascript/examples/distance-matrix">their distance service</a> to compute distances.

blocks|key|2634849|text|为源点和目标点创建两个DbGeography实例(http://msdn.microsoft.com/en-us/library/system.data.spatial.dbgeography.distance(v=vs.110).aspx然后利用DbGeography中的距离法求出两点间的距离。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2634850|有关距离单位，请参阅此帖子：System.Data.Spatial+DbGeography.Distance+units?|2634851|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/system.data.spatial.dbgeography.distance(v=vs.110%2529.aspx)|1|https://stackoverflow.com/questions/13934169/system-data-spatial-dbgeography-distance-units^0|P|2N|0|0|E|1B|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

Create two DbGeography instance for source and destination point (<a href="http://msdn.microsoft.com/en-us/library/system.data.spatial.dbgeography.distance(v=vs.110).aspx" rel="nofollow noreferrer">http://msdn.microsoft.com/en-us/library/system.data.spatial.dbgeography.distance(v=vs.110).aspx</a>). Then use distance method of DbGeography for find the distance between two points.

For distance unit refer this post : <a href="https://stackoverflow.com/questions/13934169/system-data-spatial-dbgeography-distance-units">System.Data.Spatial DbGeography.Distance units?</a>

blocks|key|4039050|text|我已经做到了这一点，但在Java语言中，也许你可以很容易地将其移植到c#。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4039051|对于小于2000公里的距离，这是非常准确的，然后它可能会与实际距离略有不同。这是由于地球的曲率造成的。但对于较小的距离，您可以假设它是简单的，对结果没有影响或影响很小。|4039052|这是我发现的一个有用的link。|offset|length|4039053|和另一个link，在C#中实现了哈维正弦。|4039054|希望能有所帮助。|4039055|祝你好运，费德里科。|4039056|++++public+String+execute(String+plat1,+String+plon1,+String+plat2,+String+plon2)+{
++++++++String+distance;
++++++++double+lat1,+lon1,+lat2,+lon2;
++++++++try{
++++++++++++lat1+=+Double.parseDouble(plat1);
++++++++++++lon1+=+Double.parseDouble(plon1);
++++++++++++lat2+=+Double.parseDouble(plat2);
++++++++++++lon2+=+Double.parseDouble(plon2);
++++++++}+catch+(Exception+e){
++++++++++++lat1+=+0.0d;
++++++++++++lon1+=+0.0d;
++++++++++++lat2+=+0.0d;
++++++++++++lon2+=+0.0d;
++++++++}

++++++++//theta+and+distance
++++++++double+theta+=+lon1+-+lon2;
++++++++double+dist+=+Math.sin(deg2rad(lat1))+*+Math.sin(deg2rad(lat2))+%2B+Math.cos(deg2rad(lat1))+*+Math.cos(deg2rad(lat2))+*+Math.cos(deg2rad(theta));

++++++++//distance
++++++++dist+=+Math.acos(dist);
++++++++dist+=+rad2deg(dist);
++++++++dist+=+dist+*+60+*+1.1515;
++++++++dist+=+dist+*+1.609344;

++++++++//convert+to+meters
++++++++dist+=+dist+*+1000;

++++++++//output+in+meters
++++++++distance+=+Double.toString(dist);
++++++++return+distance;
++++}

++++private+double+deg2rad(double+deg)+{
++++++++return+(deg+*+Math.PI+/+180.0);
++++}

++++private+double+rad2deg(double+rad)+{
++++++++return+(rad+*+180.0+/+Math.PI);
++++}|code-block|syntax|javascript|4039057|一个小小的解释：|4039058|execute|4039059|(plat1，plon1)是后面，点1或原点|unordered-list-item|4039060|
|4039061|(plat2，plon2)是后面，点2或execute方法的lng是您调用的方法，它返回一个包含以米为单位的距离的字符串(转换为其他单位很容易)|4039062|4039063|两个辅助函数被声明为"deg2rad“和”rad2deg“。|4039064|4039065|entityMap|0|LINK|mutability|MUTABLE|url|http://www.movable-type.co.uk/scripts/latlong.html|1|http://www.codecodex.com/wiki/Calculate_Distance_Between_Two_Points_on_a_Globe#C.23^0|0|0|B|4|0|0|4|4|1|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1J|8|@]|9|@$F|1K|G|1L|1|1M]]|A|$]]|$1|H|3|I|5|6|7|1N|8|@]|9|@$F|1O|G|1P|1|1Q]]|A|$]]|$1|J|3|K|5|6|7|1R|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1S|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1T|8|@]|9|@]|A|$Q|R]]|$1|S|3|T|5|6|7|1U|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1V|8|@]|9|@]|A|$]]|$1|W|3|X|5|Y|7|1W|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1X|8|@]|9|@]|A|$]]|$1|11|3|12|5|Y|7|1Y|8|@]|9|@]|A|$]]|$1|13|3|10|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|14|3|15|5|Y|7|20|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|21|8|@]|9|@]|A|$]]|$1|17|3|-4|5|6|7|22|8|@]|9|@]|A|$]]]|18|$19|$5|1A|1B|1C|A|$1D|1E]]|1F|$5|1A|1B|1C|A|$1D|1G]]]]

I've done this but in Java, probably you can port it to c# with no much effort.

This is very accurate for distances shorter than 2000 kilometers, then it can vary a little bit from the real distance. This is due to the earth curvature. But for small distances you can assume it is plain with none or very small impact in the result.

Here is a <a href="http://www.movable-type.co.uk/scripts/latlong.html" rel="nofollow">link</a> I found useful.

And another <a href="http://www.codecodex.com/wiki/Calculate_Distance_Between_Two_Points_on_a_Globe#C.23" rel="nofollow">link</a>, an implementation of Haversine in C#.

Hope it helps.

Best regards, Federico.

<pre><code> public String execute(String plat1, String plon1, String plat2, String plon2) {
 String distance;
 double lat1, lon1, lat2, lon2;
 try{
 lat1 = Double.parseDouble(plat1);
 lon1 = Double.parseDouble(plon1);
 lat2 = Double.parseDouble(plat2);
 lon2 = Double.parseDouble(plon2);
 } catch (Exception e){
 lat1 = 0.0d;
 lon1 = 0.0d;
 lat2 = 0.0d;
 lon2 = 0.0d;
 }

 //theta and distance
 double theta = lon1 - lon2;
 double dist = Math.sin(deg2rad(lat1)) * Math.sin(deg2rad(lat2)) + Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * Math.cos(deg2rad(theta));

 //distance
 dist = Math.acos(dist);
 dist = rad2deg(dist);
 dist = dist * 60 * 1.1515;
 dist = dist * 1.609344;

 //convert to meters
 dist = dist * 1000;

 //output in meters
 distance = Double.toString(dist);
 return distance;
 }

 private double deg2rad(double deg) {
 return (deg * Math.PI / 180.0);
 }

 private double rad2deg(double rad) {
 return (rad * 180.0 / Math.PI);
 }
</code></pre>

A little explanation:

<ul>
<li>(plat1, plon1) is lat and lng of point 1 or origin</li>
<li>(plat2, plon2) is lat and lng of point 2 or destination</li>
<li>the method "execute" is the one you call, it returns a string containing the distance in meters (conversion to other units can be done easily)</li>
<li>two assisting functions are declared as "deg2rad" and "rad2deg".</li>
</ul>

blocks|key|4039141|text|你可能会找到一个简单的答案，就是一个简单的公式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4039142|Calculate+distance+between+two+points+in+google+maps+V3|offset|length|4039143|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|4039144|使用google地图api将是最好的解决方案：|4039145|https://developers.google.com/maps/documentation/distancematrix/|4039146|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/1502590/calculate-distance-between-two-points-in-google-maps-v3|1^0|0|0|1J|0|0|0|0|0|1S|1|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@$D|Z|E|10|1|11]]|A|$]]|$1|F|3|G|5|H|7|12|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|14|8|@]|9|@$D|15|E|16|1|17]]|A|$]]|$1|O|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|N]]]]

The simple answer you may find , is a simple formula :

<a href="https://stackoverflow.com/questions/1502590/calculate-distance-between-two-points-in-google-maps-v3">Calculate distance between two points in google maps V3</a>

<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>

using google maps api will be the best solution:

<a href="https://developers.google.com/maps/documentation/distancematrix/" rel="nofollow noreferrer">https://developers.google.com/maps/documentation/distancematrix/</a>

I am looking to add a feature to a order entry system to show the distance in KM between the order location and assigned delivery van in real time.

The vans have a GPS Tracking system from <a href="http://highpointgps.com" rel="nofollow">High Point GPS</a>, comes with an API to query driver location, returns in format of LAT/LONG, and location address if available.

Once I get the location I can use a web service to calculate the distance between the 2 locations, does anyone know what is the best way to get the distance between 2 locations using LAT/LONG for vehicle and address lookup for the delivery location?

Vehicle GPS Tracking Devices

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我希望添加一个功能，以订单输入系统，以显示之间的距离在公里的订单位置和指定的送货货车在实时。面包车有的全球定位系统，配备了一个应用程序接口来查询司机的位置，以LAT/LONG格式返回，以及位置地址(如果可用)。一旦我得到了位置，我就可以使用web服务来计算两个位置之间的距离，有谁知道使用LAT/LONG来获取两个位置之间距离的最佳方法是什么？

问车载GPS跟踪设备
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回答 5

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