SQL Oracle 中按行进行的项目的条件计数

如何解决SQL Oracle 中按行进行的项目的条件计数

我发布了类似的问题

conditional count in a list of strings in SQL Oracle

这次数据集是按行进行的，但逻辑保持不变。这些项目将被一一添加到 list1,2,3..。我在将项目添加到列表时使用了日期。

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','toy'         FROM DUAL UNION ALL
SELECT 'list2','03-01-2020','car'         FROM DUAL UNION ALL
SELECT 'list3','toy'         FROM DUAL UNION ALL
SELECT 'list3','cards'       FROM DUAL UNION ALL
SELECT 'list3','cards'       FROM DUAL UNION ALL
SELECT 'list4','car'         FROM DUAL UNION ALL
SELECT 'list4','cards'       FROM DUAL UNION ALL
SELECT 'list5','toy'         FROM DUAL UNION ALL
SELECT 'list5','04-01-2020','cards'       FROM DUAL UNION ALL
SELECT 'list6','car'         FROM DUAL UNION ALL
SELECT 'list6','toy'         FROM DUAL UNION ALL
SELECT 'list6','cards'       FROM DUAL;

表格看起来像这样

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

解决方法

您可以使用窗口函数来获取有关列表的信息，然后可以将其用于过滤：

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;

Here 是一个 dbfiddle。