如何避免 MySQL 更新中相关子查询分配中的多列

Question

我正在尝试根据社区邮政编码并使用 Haversine formula with SQL described here 将最近的位置分配给社区。我需要 return 一个标量值，但我似乎无法避免使用确定最近位置所需的第二个计算距离值。帮助。

UPDATE Community AS c
    JOIN Postcode p on p.id = c.postcode_id
    JOIN (
        SELECT 100.0 AS radius, 111.045 AS distance_unit
    ) AS a
SET c.location_id = (
    SELECT l.id,
        a.distance_unit
            * DEGREES(ACOS(COS(RADIANS(p.latitude))
            * COS(RADIANS(l.latitude))
            * COS(RADIANS(p.longitude - l.longitude))
            + SIN(RADIANS(p.latitude))
            * SIN(RADIANS(l.latitude)))) AS distance
    FROM Location AS l
    WHERE l.latitude
        BETWEEN p.latitude  - (a.radius / a.distance_unit)
            AND p.latitude  + (a.radius / a.distance_unit)
    AND l.longitude
        BETWEEN p.longitude - (a.radius / (a.distance_unit * COS(RADIANS(p.latitude))))
            AND p.longitude + (a.radius / (a.distance_unit * COS(RADIANS(p.latitude))))
    HAVING distance <= a.radius
    ORDER BY distance
    LIMIT 1
)

Answer 1

使用您拥有的结构，您需要将距离计算移动到 WHERE 和 ORDER BY 子句中：

SET c.location_id = (
    SELECT l.id
    FROM Location AS l
    WHERE l.latitude
        BETWEEN p.latitude  - (a.radius / a.distance_unit)
            AND p.latitude  + (a.radius / a.distance_unit)
    AND l.longitude
        BETWEEN p.longitude - (a.radius / (a.distance_unit * COS(RADIANS(p.latitude))))
            AND p.longitude + (a.radius / (a.distance_unit * COS(RADIANS(p.latitude))))
    AND a.distance_unit
            * DEGREES(ACOS(COS(RADIANS(p.latitude))
            * COS(RADIANS(l.latitude))
            * COS(RADIANS(p.longitude - l.longitude))
            + SIN(RADIANS(p.latitude))
            * SIN(RADIANS(l.latitude)))) <= a.radius
    ORDER BY a.distance_unit
            * DEGREES(ACOS(COS(RADIANS(p.latitude))
            * COS(RADIANS(l.latitude))
            * COS(RADIANS(p.longitude - l.longitude))
            + SIN(RADIANS(p.latitude))
            * SIN(RADIANS(l.latitude)))) 
    LIMIT 1
)

如何避免 MySQL 更新中相关子查询分配中的多列

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