SQL Oracle中按行项目的条件计数

我发布了类似的问题

这次数据集是按行的，但逻辑保持不变。这些项目将添加到列表1、2、3中。。逐一地。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

如果列表中只有一辆车，那么你要计算汽车的数量。 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

e、 g

---

您可以使用窗口函数获取有关列表的信息，这些信息可用于筛选：

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;

---

是一个dbfiddle。

您的输出不相同，应该有3列，但无论如何，计数是错误的，与我想要的输出不同table@Pato . . . 问题很微妙。我修正了答案，加了一把小提琴。

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;