我有一个在 MySQL 上运行没有问题的子查询,我想在 AWS-Redshift 上使用相同的查询,但我收到此错误:
[0A000] ERROR: This type of correlated subquery pattern is not supported yet
,这是查询:
SELECT
COALESCE(r.id, r2.id) AS region_id,
FROM
search s
LEFT JOIN
region r ON s.region_id = r.id
LEFT JOIN
region r2 ON r2.id = (SELECT id
FROM region AS r
WHERE (acos(sin(radians(s.latitude))
* sin(radians(r.latitude))
+ cos(radians(s.latitude))
* cos(radians(r.latitude))
* cos(radians(r.longitude) - radians(s.longitude))
) * 3959 < :dis
)
AND type IN (1)
ORDER BY
(
acos
(
sin(radians(s.latitude))
* sin(radians(r.latitude))
+ cos(radians(s.latitude))
* cos(radians(r.latitude))
* cos(radians(r.longitude) - radians(s.longitude))
)
* 3959
) ASC LIMIT 1)
WHERE
s.user IS NOT NULL
ORDER BY
s.date_created DESC;
到目前为止,我发现这部分代码就是问题所在:
( acos
(
sin(radians(s.latitude))
* sin(radians(r.latitude))
+ cos(radians(s.latitude))
* cos(radians(r.latitude))
* cos(radians(r.longitude) - radians(s.longitude))
) * 3959 < :dis
)
AND type IN (1)
ORDER BY
(
acos
(
sin(radians(s.latitude))
* sin(radians(r.latitude))
+ cos(radians(s.latitude))
* cos(radians(r.latitude))
* cos(radians(r.longitude) - radians(s.longitude))
)
* 3959
)
但我不知道如何在没有子查询的情况下做到这一点。
问题在于 ON 子句中的子查询需要针对每种连接可能性重新评估。在集群数据库上,这是非常昂贵的。因此,您需要将其展平为一组附加的 JOIN 信息,其中包含连接 r 和 r2 所需的所有信息。
我可以尝试一下,但我不知道你的数据,我只是猜测查询中重要的内容。
SELECT
COALESCE(r.id, c.id2) AS region_id,
FROM
search s
LEFT JOIN
region r ON s.region_id = r.id
LEFT JOIN ( SELECT id1, id2
FROM ( SELECT s.id as id1, x.id as id2,
acos(sin(radians(t.latitude))
* sin(radians(x.latitude))
+ cos(radians(t.latitude))
* cos(radians(x.latitude))
* cos(radians(x.longitude) - radians(t.longitude))
) * 3959 as calc,
ROW_NUMBER() OVER (partition by id1 order by calc asc) as rn
FROM region AS x
CROSS JOIN search t
WHERE calc < :dis
AND type IN (1)
AND t.user IS NOT NULL)
WHERE rn = 1) c
ON c.id = s.id
WHERE
s.user IS NOT NULL
ORDER BY
s.date_created DESC;
我无法对此进行测试,因此希望此更改能为您提供一个起点。
注意交叉连接。这是相关子查询的稍微便宜的替代方案。无论哪种方式,您都会计算区域和搜索之间的每个行组合的距离(?),然后找到最小值。