备用行值减去SQL

Question

嗨大家我的查询中有一个数据集，如下所示：

Date   |  Count   | Activity
10 Nov |    10    |    A
11 Nov |    11    |    A
10 Nov |    12    |    B
11 Nov |    13    |    B

我正在努力实现这一结果。基本上逻辑是第2行将减去第1行，第4行减去第3行。

Date   |  Count   | Activity  | Diff
10 Nov |    10    |    A      |  0
11 Nov |    11    |    A      |  1
10 Nov |    12    |    B      |  0
11 Nov |    13    |    B      |  1

我当前的查询如下所示：

select  DATE, count(distinct(ID)) as Count,
count(distinct(ID)) - LAG(count(distinct(ID)),1) over (order by count(distinct(ID))) as Eng_change
from (Select DATA.*,PRODUCT.MAPPING from DATA left join PRODUCT on DATA.Part_Number=PRODUCT.PRODUCT_NUMBER ) OVERALLFUNNEL 
WHERE ACTIVITY_RANK>5
group by OVERALLFUNNEL.ACTIVITY,OVERALLFUNNEL.DATE
ORDER BY ACTIVITY_RANK ASC

使用滞后将总是减去上一行，但这不是我想要的。

任何帮助或功能？

问候

Answer 1

如果我正确理解您的问题，您希望每项活动都有所不同。如果是这样的话：

select DATE, activity,
       count(distinct ID) as Count,
       (count(distinct ID) -
        LAG(count(distinct ID), 1) over (partition by activity
                                         order by count(distinct ID)
                                        )
      ) as Eng_change

这将为NULL提供第一个值。如果您需要0，请使用coalesce()或类似内容。