我有一个数据库,该数据库可跟踪一段时间内的人员工资,如下表:
我不知道要使用什么查询,因为它需要在薪水数据库中进行迭代以检查特定日期的有效薪水。 有这个主意吗?
谢谢!
答案 0 :(得分:2)
这里包含所有示例数据。如前所述,您需要一个有效的起始日期...
WITH
-- your input ...
indata(id,datevaliduntil,salary) AS (
SELECT 1001,DATE '9999-12-31', 5000
UNION ALL SELECT 1001,DATE '2020-08-31', 4000
UNION ALL SELECT 1001,DATE '2020-04-30', 3000
)
,
-- make it almost like a slowly changing dimension
-- table - ad a valid-from-date ...
scd AS (
SELECT
id
, LAG(datevaliduntil,1,DATE '1900-01-01') OVER (
PARTITION BY id ORDER BY datevaliduntil
) AS datevalidfrom
, datevaliduntil
, salary
FROM indata
)
,
-- the months from the example ...
months(monthend) AS (
SELECT
mon::DATE - 1 AS monthend
FROM
GENERATE_SERIES(
'2020-04-01'::DATE
, '2021-03-01'::DATE
, INTERVAL '1 MONTH'
) gs(mon)
)
SELECT
monthend
, id
, salary
FROM scd
JOIN months ON monthend > datevalidfrom
AND monthend <= datevaliduntil
ORDER BY 1
;
-- out monthend | id | salary
-- out ------------+------+--------
-- out 2020-03-31 | 1001 | 3000
-- out 2020-04-30 | 1001 | 3000
-- out 2020-05-31 | 1001 | 4000
-- out 2020-06-30 | 1001 | 4000
-- out 2020-07-31 | 1001 | 4000
-- out 2020-08-31 | 1001 | 4000
-- out 2020-09-30 | 1001 | 5000
-- out 2020-10-31 | 1001 | 5000
-- out 2020-11-30 | 1001 | 5000
-- out 2020-12-31 | 1001 | 5000
-- out 2021-01-31 | 1001 | 5000
-- out 2021-02-28 | 1001 | 5000
答案 1 :(得分:1)
这是使用横向连接的便利位置。以下是在每月的第一天而不是最后一天-因为这样更易于生成:
select i.id, gs.mon, s.salary
from generate_series('2019-01-01'::date, '2020-12-01'::date, interval '1 month') gs(mon) cross join
(select distinct id from salaries) i left join lateral
(select s.salary
from salaries s
where s.id = i.id and s.datevaliduntil >= gs.mon
order by s.datevaliduntil asc
limit 1
) s;
当然,如果需要最后一天,您可以从每个日期中减去1天。
答案 2 :(得分:1)
我将使用横向联接,但反之亦然:从表本身开始,将上一个日期带入lag()
,然后使用generate series生成之间的日期。需要一些额外的逻辑来调整月底:
select x.date - interval '1 day' date, t.id, t.salary
from (
select id, salary,
datevaliduntil + interval '1 day' datevaliduntil,
lag(datevaliduntil, 1, datevaliduntil)
over(partition by id order by datevaliduntil) + interval '1 day' lag_datevaliduntil
from mytable t
) t
cross join lateral generate_series(
t.lag_datevaliduntil,
least(t.datevaliduntil, '2021-03-01'),
'1 month'
) x(date)
您可以在generate_series
的第二个参数中使用文字日期控制总体上限(此处,您要停止2021年3月底)。
date | id | salary :------------------ | ---: | -----: 2020-04-30 00:00:00 | 1001 | 3000 2020-04-30 00:00:00 | 1001 | 4000 2020-05-31 00:00:00 | 1001 | 4000 2020-06-30 00:00:00 | 1001 | 4000 2020-07-31 00:00:00 | 1001 | 4000 2020-08-31 00:00:00 | 1001 | 4000 2020-08-31 00:00:00 | 1001 | 5000 2020-09-30 00:00:00 | 1001 | 5000 2020-10-31 00:00:00 | 1001 | 5000 2020-11-30 00:00:00 | 1001 | 5000 2020-12-31 00:00:00 | 1001 | 5000 2021-01-31 00:00:00 | 1001 | 5000 2021-02-28 00:00:00 | 1001 | 5000