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我发布了类似的问题

SQL Oracle中字符串列表中的条件计数

这次数据集是按行排列的，但逻辑保持不变。这些项目将被添加到 list1,2,3.. 一个一个。我使用了将项目添加到列表中的日期。

这是数据

CREATE TABLE lists ( column1, column2 , column3) AS
SELECT 'list1', '01-01-2020' , 'car'         FROM DUAL UNION ALL
SELECT 'list1', '02-01-2020' , 'car'         FROM DUAL UNION ALL
SELECT 'list2', '01-01-2020' , 'car'         FROM DUAL UNION ALL
SELECT 'list2', '02-01-2020' , 'toy'         FROM DUAL UNION ALL
SELECT 'list2', '03-01-2020' , 'car'         FROM DUAL UNION ALL
SELECT 'list3', '01-01-2020' , 'toy'         FROM DUAL UNION ALL
SELECT 'list3', '02-01-2020' , 'cards'       FROM DUAL UNION ALL
SELECT 'list3', '03-01-2020' , 'cards'       FROM DUAL UNION ALL
SELECT 'list4', '01-01-2020' , 'car'         FROM DUAL UNION ALL
SELECT 'list4', '02-01-2020' , 'cards'       FROM DUAL UNION ALL
SELECT 'list4', '03-01-2020' , 'cards'       FROM DUAL UNION ALL
SELECT 'list5', '01-01-2020' , 'toy'         FROM DUAL UNION ALL
SELECT 'list5', '02-01-2020' , 'cards'       FROM DUAL UNION ALL
SELECT 'list5', '03-01-2020' , 'toy'         FROM DUAL UNION ALL
SELECT 'list5', '04-01-2020' , 'cards'       FROM DUAL UNION ALL
SELECT 'list6', '01-01-2020' , 'car'         FROM DUAL UNION ALL
SELECT 'list6', '02-01-2020' , 'cards'       FROM DUAL UNION ALL
SELECT 'list6', '03-01-2020' , 'toy'         FROM DUAL UNION ALL
SELECT 'list6', '04-01-2020' , 'cards'       FROM DUAL;

桌子看起来像这样

COLUMN1 COLUMN2 COLUMN3
list1   01-01-2020  car
list1   02-01-2020  car
list2   01-01-2020  car
list2   02-01-2020  toy
list2   03-01-2020  car
list3   01-01-2020  toy
list3   02-01-2020  cards
list3   03-01-2020  cards
list4   01-01-2020  car
list4   02-01-2020  cards
list4   03-01-2020  cards
list5   01-01-2020  toy
list5   02-01-2020  cards
list5   03-01-2020  toy
list5   04-01-2020  cards
list6   01-01-2020  car
list6   02-01-2020  cards
list6   03-01-2020  toy
list6   04-01-2020  cards

如果列表中只有 CAR，那么您计算汽车的数量。如果有非汽车项目，则计算添加到列表中的第一个非汽车项目

例如

list1  car   2
list2  toy   1
list3  toy   1
list4  cards 2
list5  toy   2
list6  cards 2
4

1 回答 1

1

您可以使用窗口函数来获取有关列表的信息，然后可用于过滤：

select l.column1, item_to_count,
       count(*)
from (select l.*,
             min(l.column2) keep (dense_rank first order by (case when l.column3 <> 'car' then 1 else 2 end), l.column2) over (partition by l.column1) as date_to_count,
             min(l.column3) keep (dense_rank first order by (case when l.column3 <> 'car' then 1 else 2 end), l.column2) over (partition by l.column1) as item_to_count
      from lists l
     ) l
where column3 = item_to_count and column2 >= date_to_count
group by l.column1, item_to_count
order by l.column1;

这是一个 db<>fiddle。

于 2021-01-12T18:25:33.250 回答