我有2个表格,如下面的示例,towns
和things
,并且需要获取距离该城镇记录x
距离附近的城镇附近的城镇列表。纬度和经度将用于进行距离计算。
我已经查看了其他一些问题,并设法从things
获取单个指定城镇的记录,但无法想到如何获取附近things
所有城镇的列表离它们的距离比x
更近。
能够根据x
距离内附近物品的数量对结果城镇进行排序将是一个奖励。
TOWNS
+--------+----------+---------+---------+
| townId | townName | townLat | townLng |
+--------+----------+---------+---------+
| 1 | town a | 1.5 | 1.9 |
| 2 | town b | 1.4 | 3.8 |
| 3 | town c | 2.3 | 2.7 |
| 4 | town d | 3.2 | 1.6 |
| ... | ... | ... | ... |
+--------+----------+---------+---------+
THINGS
+---------+-----------+----------+----------+
| thingId | thingName | thingLat | thingLng |
+---------+-----------+----------+----------+
| 1 | thing a | 2.1 | 3.1 |
| 2 | thing b | 1.1 | 2.3 |
| 3 | thing c | 3.2 | 0.2 |
| 4 | thing d | 1.3 | 1.1 |
| ... | ... | ... | ... |
+---------+-----------+----------+----------+
提前致谢
答案 0 :(得分:2)
您可以CROSS JOIN
获取城镇和事物的所有可能组合,然后计算每个城镇和物体之间的Haversine距离。我使用SELECT DISTINCT
来确保城镇只在结果集中列出一次。
SELECT DISTINCT TOWNS.townName FROM
TOWNS CROSS JOIN THINGS
WHERE 3959 * acos(
cos(radians( TOWNS.townLat ))
* cos(radians( THINGS.thingLat ))
* cos(radians( TOWNS.townLng ) - radians( THINGS.thingLng ))
+ sin(radians( TOWNS.townLat ))
* sin(radians( THINGS.thingLat ))
) < x
我使用的公式为英里x
(地球的平均半径为3959英里)。