我有一个具有闲置结构的表account
:
| agg_type | agg_id | sequence | payload | is_snapshot | timestamp |
| "account" | "agg_1" | 1 | "..." | false | ... |
| "account" | "agg_1" | 2 | "..." | true | ... |
| "account" | "agg_1" | 3 | "..." | false | ... |
| "account" | "agg_1" | 4 | "..." | false | ... |
| "account" | "agg_1" | 5 | "..." | false | ... |
| "account" | "agg_1" | 6 | "..." | false | ... |
| "account" | "agg_1" | 7 | "..." | true | ... |
| "account" | "agg_1" | 8 | "..." | false | ... |
我需要编写一个查询,从特定聚合的最新快照开始,从该表中检索所有行。例如,在这个表的情况下,查询将返回最后两行(序列7和8(。
我认为查询会像一样
SELECT * FROM account
WHERE
agg_type='account'
AND agg_id='agg_1'
ORDER BY sequence ASC
LIMIT (???);
我不太确定如何实现(???)
部分。
Obs:
- 如果有任何帮助的话,我会使用Postgres
- (agg_type,agg_id,sequence(组合是主键
简单地说,我们可以检索序列大于或等于快照的最高序列id的所有帐户
SELECT * FROM account a
WHERE
a.agg_type='account'
AND a.agg_id='agg_1'
AND a.sequence >=
(SELECT MAX(sequence) FROM account b WHERE a.agg_type = b.agg_type AND a.agg_id = b. agg_id AND b.is_snapshot = true)
如果你想做所有的事情,把它写成一个连接可能会更清楚:
SELECT a.*
FROM
account a
INNER JOIN
(
SELECT
agg_type,
agg_id,
MAX(sequence) as maxseq
FROM account b
GROUP BY agg_type, add_id
) maxes
ON
a.agg_type = maxes.agg_type and
maxes.agg_id = a.max_id and
a.sequence >= maxes.maxseq
这并不是说我们不能用任何一种形式完成任何一项任务(内部postgres可能无论如何都会执行相同的任务(,但我一直觉得使用联接作为"的限制;这里有10000行,我只想要满足这1000行所规定的标准的2000行;最清楚地认为是结合在一起的数据块
使用AS(选择*,row_number((over(partition BY a.agg_type,a.agg_id ORDER BY a."SEQUENCE"DESC(rnkFROM帐户a)从WHERE选择*a.rnk<=2.
窗口函数只需一种排序即可为所有(agg_type, agg_id)
组合提取此项:
with mark as (
select *,
bool_or(is_snapshot) over w as trail_true
from account
window w as (partition by agg_type, agg_id
order by sequence
rows between 1 following
and unbounded following)
)
select *
from mark
where not coalesce(trail_true, false)
order by agg_type, agg_id, sequence