Question

我有一张桌子，里面有以下数据。

Batch      Ingredient       Transaction date
----   -----------------  --------------------
A              I1              2019/05/22
A              I2              2019/05/23
A              I3              2019/05/24

我想按以下格式按交易日期显示最多两个记录

Batch      Ingredient1     TransDate1    Ingredient2     TransacDate2
----    ---------------- ------------- ---------------- --------------    
A              I3          2019/05/24        I2           2019/05/23

成分1应该是最新记录，并且成分是前一个记录。没有成分3。

我尝试了数据透视表，但是没有用。

Answer 1

您可以使用row_number()和条件聚合：

select batch,
       max(case when seqnum = 1 then ingredient end) as ingredient1,
       max(case when seqnum = 1 then transdate end) as transdate1,
       max(case when seqnum = 2 then ingredient end) as ingredient2,
       max(case when seqnum = 2 then transdate end) as transdate2
from (select t.*,
             row_number() over (partition by batch order by transdate desc) as seqnum
      from t
     ) t
where seqnum <= 2
group by batch;

如何将两行转换为一行作为先行记录和当前记录

1 个答案: