SQL - 如何仅获取每个 ID 的当前值和先前值

Question

我有一张 table

id, date, value
a,  5/22/2019, 22
b,  5/22/2019, 22
c,  5/22/2019, 22
a,  5/21/2019, 21
b,  5/21/2019, 21
c,  5/21/2019, 21
a,  5/20/2019, 20
b,  5/20/2019, 20
c,  5/20/2019, 20

想要获取当前(最新)和之前日期的 ID 和值:

id, date, date-1
a, 22, 21
b, 22, 21
c, 22, 21

到目前为止，我有一些部分，但每行只需要 1 个 ID(下面为每个 ID 提供了超过 1 行 - 对于相同的 ID，它将在下一行显示 5/20 日期，这是错误的):

select 
    id, value,
    lag(value, 1, 0) over (partition by id order by date ) as "date - 1"
from 
    table1

如何实现？

Answer 1

你可以像下面这样尝试

with cte as
(
select 
    id, value,
    lag(value, 1, 0) over (partition by id order by date ) as "date - 1",
    lag(value, 2, 0) over (partition by id order by date ) as "date - 2",
    row_number() over(partition by id order by date desc) rn
from 
    table1
) select * from cte where rn=1