Mysql SQL:在给定纬度的不同半径内查找最近点&;经度
到目前为止,我已经能够在同一半径内找到最近的点(本文中的仓库)。(已经有很多答案的常见问题) 这是实际代码(它不使用多个半径) 模式:Mysql SQL:在给定纬度的不同半径内查找最近点&;经度,mysql,sql,wordpress,Mysql,Sql,Wordpress,到目前为止,我已经能够在同一半径内找到最近的点(本文中的仓库)。(已经有很多答案的常见问题) 这是实际代码(它不使用多个半径) 模式: meta_id | post_id | meta_key | meta_value ------------------------------------------------------ 324802 | 9714 | warehouse_latitude | 47.1978754 324809 | 9715 |
meta_id | post_id | meta_key | meta_value
------------------------------------------------------
324802 | 9714 | warehouse_latitude | 47.1978754
324809 | 9715 | warehouse_latitude | 47.2064462
324814 | 9716 | warehouse_latitude | 47.214434
324803 | 9714 | warehouse_longitude | -1.54441
324810 | 9715 | warehouse_longitude | -1.5461347
324815 | 9716 | warehouse_longitude | -1.565993
324806 | 9714 | warehouse_radius | 5
324811 | 9715 | warehouse_radius | 2
324816 | 9716 | warehouse_radius | 10
因此,这里是解决方案(根据上一个查询更新)
如果您有更有效/可读的方法来计算距离,请不要介意给另一个anwser
SELECT DISTINCT
warehouse_latitude.post_id,
warehouse_longitude.meta_value as longitude,
warehouse_latitude.meta_value as latitude,
warehouse_radius.meta_value as radius,
((ACOS(SIN($latitude * PI() / 180) * SIN(warehouse_latitude.meta_value * PI() / 180) + COS($latitude * PI() / 180) * COS(warehouse_latitude.meta_value * PI() / 180) * COS(($longitude - warehouse_longitude.meta_value) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS distance
FROM {$this->wpdb->postmeta} as warehouse_latitude
LEFT JOIN {$this->wpdb->postmeta} as warehouse_longitude ON warehouse_latitude.post_id = warehouse_longitude.post_id
LEFT JOIN {$this->wpdb->postmeta} as warehouse_radius ON warehouse_latitude.post_id = warehouse_radius.post_id
WHERE warehouse_latitude.meta_key = 'warehouse_latitude' AND warehouse_longitude.meta_key = 'warehouse_longitude' AND warehouse_radius.meta_key = 'warehouse_radius'
HAVING distance < (radius * 0.62)
ORDER BY distance ASC
LIMIT 1
WHERE语句中的另一个条件:
AND warehouse_radius.meta_key = 'warehouse_radius'
因此:
我可以将HAVING语句更新为:
HAVING distance < (radius * 0.62) -- Km to Miles
距离<(半径*0.62)--公里到英里
不是答案。评论太长了
1 LEFT JOIN warehouse_longitude ON ... -- This is an INNER JOIN because of LINE 3
2 LEFT JOIN warehouse_radius ON ... -- This is also an INNER JOIN, because of LINE 4
3 WHERE warehouse_longitude.meta_key = 'warehouse_longitude'
4 AND warehouse_radius.meta_key = 'warehouse_radius'
因此,您不妨先编写这些内部联接。能否以表格的形式添加示例数据,然后显示预期的输出?这是一个标准的Haversine MySQL查询,但我看不出您在这里想要什么。。。其中x=…与
内部联接x
相同。因此,将条件移到on子句。还有,弧度有什么问题吗?这里有什么问题吗?@草莓当然,我不介意使用弧度,这是我找到的第一个解决方案。@SalmanA有什么问题吗?它可以不受限制地工作,但我只需要得到一个:)。尝试使用具有距离<(warehouse\u radius.radius*0.62)
确保我可以做到这一点。实际上,我在返回的响应中使用了这个别名(因为我以后可能会使用它;)。同样。谢谢你的意见。我将尝试并更新我的答案:)。
HAVING distance < (radius * 0.62) -- Km to Miles
1 LEFT JOIN warehouse_longitude ON ... -- This is an INNER JOIN because of LINE 3
2 LEFT JOIN warehouse_radius ON ... -- This is also an INNER JOIN, because of LINE 4
3 WHERE warehouse_longitude.meta_key = 'warehouse_longitude'
4 AND warehouse_radius.meta_key = 'warehouse_radius'