2 个日期之间的天数取决于 ID

Question

我有 ID 和订单日期列；我想要做的是为每个 ID 显示下一个订单日期和当前日期之间的差异。

表格

ID	Order Date
1	01/12/2020
2	15/09/2020
2	01/10/2020
3	31/12/2020
4	01/10/2020
4	03/10/2020
4	01/12/2020
5	31/01/2020

我期待什么：

ID	Order Date	Days between
1	01/12/2020	n/a
2	15/09/2020	77 days
2	01/10/2020	n/a
3	31/12/2020	n/a
4	01/10/2020	2 days
4	03/10/2020	59 days
4	01/12/2020	n/a
5	31/01/2020	0

欢迎大家帮忙...我对 SQL 不熟悉，所以只有基本的了解

Answer 1

首先，在 SQL 中，您应该只有一个包含天数的列——一个数字。不要将值转换为字符串。你所谓的'N/A' 应该是NULL 值。

要做你想做的事，请使用lag()。是这样的：

select t.*,
       (order_date -
        lag(order_date) over (partition by id order by order_date)
       ) as date_diff
from t;

请注意，并非所有数据库都支持 - 日期。日期操作的确切语法通常因数据库而异。

Answer 2

77 之间的天数似乎是错误的。请参阅下面的结果。这与上述解决方案相同，仅使用 datediff() 代替。

create table T (ID int, OrderDate datetime)

insert into T (ID, OrderDate)
  values ( 1, cast('01/12/2020' as datetime) ),
         ( 2, cast('15/09/2020' as datetime) ),
         ( 2, cast('01/10/2020' as datetime) ),
         ( 3, cast('31/12/2020' as datetime) ),
         ( 4, cast('01/10/2020' as datetime) ),
         ( 4, cast('03/10/2020' as datetime) ),
         ( 4, cast('01/12/2020' as datetime) ),
         ( 5, cast('31/01/2020' as datetime) )

select  *, datediff(day, OrderDate, lag(orderdate) over(partition by ID order by ID, Orderdate desc)) as [Days Between]
from    T 
order by ID, OrderDate asc

ID	Orderdate	Days Between
1	2020-12-01 00:00:00.000	NULL
2	2020-09-15 00:00:00.000	16
2	2020-10-01 00:00:00.000	NULL
3	2020-12-31 00:00:00.000	NULL
4	2020-10-01 00:00:00.000	2
4	2020-10-03 00:00:00.000	59
4	2020-12-01 00:00:00.000	NULL
5	2020-01-31 00:00:00.000	NULL