SQL：从给定的纬度和经度中查找不同半径内的最近点

Question

到目前为止，我已经能够找到相同半径内的最近点（在此情况下为仓库）。 （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）

$radius = 5; // miles

$q = "
    SELECT DISTINCT 
      warehouse_latitude.post_id,
      warehouse_longitude.meta_value as longitude,
      warehouse_latitude.meta_value as latitude,
      ((ACOS(SIN($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 
    WHERE warehouse_latitude.meta_key = 'warehouse_latitude' AND warehouse_longitude.meta_key = 'warehouse_longitude'
    HAVING distance < $radius
    ORDER BY distance ASC
    LIMIT 1
";

相反：

我想选择最近的点，但是为每个点定义了半径（仓库）。

• A点（9714）：半径5英里
• B点（9715）：半径2英里
• C点（9716）：半径10英里

因此，如果点B和C在范围内，并且点B最接近，则应选择点B。

wp_postmeta表：

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

架构：

Answer 1

不是答案。评论太久了...

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，您最好将它们写为INNER JOIN。

Answer 2

因此，这里是解决方案（从上一个查询更新）。

如果您有更有效/更易理解的方式来计算距离，则不要介意再给您答案。

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

我在半径值上添加了一个连接：

LEFT JOIN {$this->wpdb->postmeta} as warehouse_radius ON warehouse_latitude.post_id = warehouse_radius.post_id 

还有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