SQLite 运行 每日计算的产品交易总和
SQLite Running sum of transactions over products calculated daily
我有一个 table（储藏室）用作分类帐，其中仅在交易发生时随时间记录产品的添加和删除。我想每天获取曾经记录的所有产品的 运行 总和。可能有更好的方法来获得此结果，但我不想旋转 table.
我想要得到的结果是这里 table 中的结果。非常感谢！
您可以 CROSS 将递归 CTE 连接到 table 的不同产品，然后将 LEFT 连接到 table 以获得 运行 用 SUM() window 函数求和：
with dates as (
select min(date) date from pantry
union all
select date(date, '+1 day')
from dates
where date < (select max(date) from pantry)
)
select *
from (
select d.date, p.produce,
sum(t.transactions) over (partition by p.produce order by d.date) running_sum
from dates d
cross join (select distinct produce from pantry) p
left join pantry t on t.date = d.date and t.produce = p.produce
)
where running_sum is not null
order by date
参见demo。
结果：
> date | produce | running_sum
> :--------- | :------ | ----------:
> 2020-01-01 | banana | 50
> 2020-01-02 | banana | 50
> 2020-01-03 | apple | 5
> 2020-01-03 | banana | 40
> 2020-01-04 | apple | 5
> 2020-01-04 | banana | 40
> 2020-01-05 | apple | 5
> 2020-01-05 | banana | 40
> 2020-01-06 | apple | 5
> 2020-01-06 | banana | 40
> 2020-01-07 | apple | 3
> 2020-01-07 | banana | 20
> 2020-01-07 | grapes | 100
> 2020-01-08 | apple | 3
> 2020-01-08 | banana | 20
> 2020-01-08 | grapes | 100
> 2020-01-09 | apple | 3
> 2020-01-09 | banana | 20
> 2020-01-09 | grapes | 50
> 2020-01-10 | apple | 3
> 2020-01-10 | banana | 20
> 2020-01-10 | grapes | 50
> 2020-01-11 | apple | -1
> 2020-01-11 | banana | 20
> 2020-01-11 | grapes | 50
> 2020-01-11 | melon | 2
我有一个 table（储藏室）用作分类帐，其中仅在交易发生时随时间记录产品的添加和删除。我想每天获取曾经记录的所有产品的 运行 总和。可能有更好的方法来获得此结果，但我不想旋转 table.
我想要得到的结果是这里 table 中的结果。非常感谢！
您可以 CROSS 将递归 CTE 连接到 table 的不同产品，然后将 LEFT 连接到 table 以获得 运行 用 SUM() window 函数求和：
with dates as (
select min(date) date from pantry
union all
select date(date, '+1 day')
from dates
where date < (select max(date) from pantry)
)
select *
from (
select d.date, p.produce,
sum(t.transactions) over (partition by p.produce order by d.date) running_sum
from dates d
cross join (select distinct produce from pantry) p
left join pantry t on t.date = d.date and t.produce = p.produce
)
where running_sum is not null
order by date
参见demo。
结果：
> date | produce | running_sum
> :--------- | :------ | ----------:
> 2020-01-01 | banana | 50
> 2020-01-02 | banana | 50
> 2020-01-03 | apple | 5
> 2020-01-03 | banana | 40
> 2020-01-04 | apple | 5
> 2020-01-04 | banana | 40
> 2020-01-05 | apple | 5
> 2020-01-05 | banana | 40
> 2020-01-06 | apple | 5
> 2020-01-06 | banana | 40
> 2020-01-07 | apple | 3
> 2020-01-07 | banana | 20
> 2020-01-07 | grapes | 100
> 2020-01-08 | apple | 3
> 2020-01-08 | banana | 20
> 2020-01-08 | grapes | 100
> 2020-01-09 | apple | 3
> 2020-01-09 | banana | 20
> 2020-01-09 | grapes | 50
> 2020-01-10 | apple | 3
> 2020-01-10 | banana | 20
> 2020-01-10 | grapes | 50
> 2020-01-11 | apple | -1
> 2020-01-11 | banana | 20
> 2020-01-11 | grapes | 50
> 2020-01-11 | melon | 2