Sql 基于字段值限制查询结果
我有一个具有休闲结构的表帐户:Sql 基于字段值限制查询结果,sql,postgresql,Sql,Postgresql,我有一个具有休闲结构的表帐户: | agg_type | agg_id | sequence | payload | is_snapshot | timestamp | | "account" | "agg_1" | 1 | "..." | false | ... | | "account" | "agg_1" | 2 | ".
| 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检索此表中的所有行。例如,对于这个表,查询将返回最后两行序列7和8
我认为这个问题会是这样的
从帐户中选择*
哪里
agg_type='account'
和agg_id='agg_1'
按顺序排序ASC
限度
是那个???我不太确定如何实施的部分
Obs:
如果有帮助的话,我正在使用Postgres。
agg_类型、agg_id、序列组合是主键。
简单地说,我们可以检索序列大于或等于最高序列id(快照)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)
如果您想完成所有这些操作,那么将其写为连接可能会更清楚:
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个表单执行任何一个任务,内部postgres可能会执行相同的任务,但我一直觉得使用连接作为限制,这里有10000行,我只想让2000行符合这1000行所规定的标准,这1000行最清楚地被认为是用AS连接在一起的数据块
挑选*
,按a.agg\u类型划分的行数,按a.SEQUENCE DESC rnk划分的a.agg\u id顺序
来自帐户a
从a.rnk中选择*窗口函数可以为所有agg_类型、agg_id组合使用一种排序:
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
您是为所有AGG还是每次为特定AGG执行此操作?每次为每个AGG执行此操作。