如何比较Oracle中一组行的最大和最小日期的列值？

Question

给定一个看起来像这样的 table，我将如何创建另一个获取此数据的查询，比较 sales_net_amt 的最早日期和最晚日期 store_id，并且 returns 每行 store_id 显示纯文本是否在另一列中“增加”或“减少”加班时间？

对于计算列，我假设我可以只使用“case when”子句，但我不确定如何导出最新日期和最早日期的两个单独值来进行比较。我能做的最好的事情就是创建一个显示最新和最早日期及其 sales_net_amt 的查询，但我不知道如何将计算列添加到现有值

的基础上

这是我进行的查询（不需要 QUARTER 列并且使查询不必要地冗长，只是想暂时添加它以防我最终使用它）：

select case when extract(month from full_date) between 1 and 3 then substr(item_scan_timestamp, -4) || 'Q1'
     when extract(month from full_date) between 4 and 6 then substr(item_scan_timestamp, -4) || 'Q2'
     when extract(month from full_date) between 7 and 9 then substr(item_scan_timestamp, -4) || 'Q3'
     when extract(month from full_date) between 10 and 12 then substr(item_scan_timestamp, -4) || 'Q4' end quarter, r.store_id, p.full_date, r.sales_net_amt
from retailsalesfact r join purchasedate p on p.purchase_date_id = r.purchase_date_id
where (r.store_id, p.full_date) in (select r.store_id, max(p.full_date) full_date
        from retailsalesfact r join purchasedate p on p.purchase_date_id = r.purchase_date_id
        group by r.store_id)
union all
select case when extract(month from full_date) between 1 and 3 then substr(item_scan_timestamp, -4) || 'Q1'
     when extract(month from full_date) between 4 and 6 then substr(item_scan_timestamp, -4) || 'Q2'
     when extract(month from full_date) between 7 and 9 then substr(item_scan_timestamp, -4) || 'Q3'
     when extract(month from full_date) between 10 and 12 then substr(item_scan_timestamp, -4) || 'Q4' end quarter, r.store_id, p.full_date, r.sales_net_amt
from retailsalesfact r join purchasedate p on p.purchase_date_id = r.purchase_date_id
where (r.store_id, p.full_date) in (select r.store_id, min(p.full_date) full_date
        from retailsalesfact r join purchasedate p on p.purchase_date_id = r.purchase_date_id
        group by r.store_id)
order by store_id, full_date, sales_net_amt;

select store_id, full_date, sales_net_amt
from retailsalesfact r join purchasedate p on p.purchase_date_id = r.purchase_date_id
order by store_id;

感谢任何建议。

Answer 1

假设您的 FULL_DATE 列实际上是一个日期，您可以使用 FIRST_VALUE and LAST_VALUE 按日期获取第一个和最后一个值，然后比较它们：

SELECT DISTINCT STORE_ID,
       CASE WHEN LAST_VALUE(SALES_NET_AMT) OVER (PARTITION BY STORE_ID ORDER BY FULL_DATE) >
                 FIRST_VALUE(SALES_NET_AMT) OVER (PARTITION BY STORE_ID ORDER BY FULL_DATE) THEN 'Increased'
            ELSE 'Decreased'
       END AS SALES_CHANGE
FROM YourTable

Answer 2

您可以在 GROUP BY 中使用 KEEP 子句，因为您希望每个商店一条记录如下：

select store_id, 
       case when max(SALES_NET_AMT) keep (dense_rank first order by FULL_DATE desc) 
                  > max(SALES_NET_AMT) keep (dense_rank first order by FULL_DATE) 
            then 'Increased' 
            else 'Decreased' 
      end as result
  from your_Table t
group by store_id

如何比较Oracle中一组行的最大和最小日期的列值？

How to compare the column values of the max and min dates of a group of rows in Oracle?

sql

oracle

group-by

case