测量两个Lat/Lng点之间的距离

Question

我在尝试通过SQL语句将最近的位置返回给用户时遇到问题.为了测试我使用完全相同的坐标,这里是我所拥有的:

SQL:

SELECT  `companies`.`customerName` , 
    (3959 * 
        ACOS(
            COS(
                RADIANS(37.367485) * COS(RADIANS(`locations`.`gps_lat`)) * 
                COS(
                    RADIANS(`locations`.`gps_lng`) - RADIANS(-77.399994) + 
                    SIN(RADIANS(37.367485)) * SIN(RADIANS(`locations`.`gps_lat`))
                )
            )
        )
    )
AS  `distance` 
FROM  `locations` 
JOIN  `companies` ON  `locations`.`co_id` 
HAVING  `distance` > 25
ORDER BY distance
LIMIT 0 , 10

结果:

| customerName  | distance           |
| SOME COMPANY  | 1914.41747964854   |

locations 表值:

gps_lat   | gps_lng
37.367485 | -77.399994

我使用了谷歌的例子,我检查了几次公式,我似乎无法想出我出错的地方.任何帮助表示赞赏.

---

编辑:
因为我知道自己在做什么似乎有些困惑:

的>被取代的,得到的结果,1914很明显大于25.

这个应用程序是将用户坐标从Android应用程序传递到我们的Web服务器.纬度和经度将从$_GET值中提取,并从我们的Web服务器MYSQL数据库中的公司进行交叉引用.

上面的语法只是我在Mysql Workbench中检查我的SQL语句.显然我正在寻找零值结果.

Answer 1

我想我已经解决了这个问题,从Googles的例子改为"一个被采用的yooper的冒险"提供的公式.使用他/她的Haversine Formula版本.
注意:上面的Google示例也使用了Haversine.

SQL:

SELECT  `companies`.`customerName` , 
(2 * (3959 * ATAN2(
          SQRT(
            POWER(SIN((RADIANS(37.367485 - `locations`.`gps_lat` ) ) / 2 ), 2 ) +
            COS(RADIANS(`locations`.`gps_lat`)) *
            COS(RADIANS(37.367485 )) *
            POWER(SIN((RADIANS(-77.399994 - `locations`.`gps_lng` ) ) / 2 ), 2 )
          ),
          SQRT(1-(
            POWER(SIN((RADIANS(37.367485 - `locations`.`gps_lat` ) ) / 2 ), 2 ) +
            COS(RADIANS(`locations`.`gps_lat`)) *
            COS(RADIANS(37.367485)) *
            POWER(SIN((RADIANS(-77.399994 - `locations`.`gps_lng` ) ) / 2 ), 2 )
          ))
        )
      ))
AS 'distance'
FROM  `locations` 
JOIN  `companies` ON  `locations`.`co_id` 
HAVING  distance < 25
ORDER BY distance
LIMIT 0 , 10

对于那些对我的应用程序感兴趣的人,最终PHP:

获取价值: ?lat=37.367485&lng=-77.399994

$earth_radius_miles = 3959; // Earth radius in miles
$earth_radius_kilometers = 6371; // Earth radius in kilometers
$result_radius = 10000; // Maximum distance in either miles or kilometers

$get_lat = $_GET["lat"];
$get_lng = $_GET["lng"];

$dbSelect = mysql_query("SELECT  `companies`.`customerName`,
(2 * (".$earth_radius_miles." * ATAN2(
          SQRT(
            POWER(SIN((RADIANS(".$get_lat." - `locations`.`gps_lat` ) ) / 2 ), 2 ) +
            COS(RADIANS(`locations`.`gps_lat`)) *
            COS(RADIANS(".$get_lat.")) *
            POWER(SIN((RADIANS(".$get_lng." - `locations`.`gps_lng` ) ) / 2 ), 2 )
          ),
          SQRT(1-(
            POWER(SIN((RADIANS(".$get_lat." - `locations`.`gps_lat` ) ) / 2 ), 2 ) +
            COS(RADIANS(`locations`.`gps_lat`)) *
            COS(RADIANS(".$get_lat.")) *
            POWER(SIN((RADIANS(".$get_lng." - `locations`.`gps_lng` ) ) / 2 ), 2 )
          ))
        )
      ))
AS 'distance'
FROM  `locations` 
JOIN  `companies` ON  `locations`.`co_id` 
HAVING  distance < ".$result_radius."
ORDER BY distance
LIMIT 0 , 10") 
or die(mysql_error());

结果:

[{"customerName":"SOME COMPANY","distance":"0"}]

我用其他坐标测试了这些结果,看起来效果很好.没有测试两点之间距离很远的东西,但我的申请并没有要求我这样做.