Question

我有一张桌子

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）和上一个日期的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，但每行仅需要1个ID（此波纹管为每个ID提供多于1行-对于同一ID，它将在下一行显示5/20日期，这是错误的）：< / p>

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

您可以尝试以下操作

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

Answer 2

这个怎么样？请注意，我将列命名为XMLHttpRequest而不是日期：

ddate

SQL-如何仅获取每个ID的当前值和先前值

2 个答案: