Question

我怀疑我需要某种窗口功能才能做到这一点。我将以下项目数据作为示例：

count | date
------+-----------
3     | 2017-09-15
9     | 2017-09-18
2     | 2017-09-19
6     | 2017-09-20
3     | 2017-09-21

首先我的数据存在空白，我在这里有另一个查询：

select until_date, until_date - (lag(until_date)  over ()) as delta_days from ranges

我已经生成了以下数据：

until_date | delta_days
-----------+-----------
2017-09-08 |
2017-09-11 | 3
2017-09-13 | 2
2017-09-18 | 5
2017-09-21 | 3
2017-09-22 | 1

所以我希望我的最终查询产生这个结果：

start_date | ending_date | total_items
-----------+-------------+--------------
2017-09-08 | 2017-09-10  | 0
2017-09-11 | 2017-09-12  | 0
2017-09-13 | 2017-09-17  | 3
2017-09-18 | 2017-09-20  | 15
2017-09-21 | 2017-09-22  | 3

根据第二个表中的自定义范围，它告诉我第一个表中每天的项目总数。

在这个特定的例子中，我将在开始和结束之间总计total_items（因为日期会有重叠，我会从结束日期减去1而不计算重复项）

任何人都知道怎么做？

谢谢！

Answer 1

使用daterange类型。请注意，您无需计算delta_days，只需将ranges转换为dataranges并使用运算符<@ - element is contained by.

with counts(count, date) as (
values
    (3, '2017-09-15'::date),
    (9, '2017-09-18'),
    (2, '2017-09-19'),
    (6, '2017-09-20'),
    (3, '2017-09-21')
),
ranges (until_date) as (
values
    ('2017-09-08'::date),
    ('2017-09-11'),
    ('2017-09-13'),
    ('2017-09-18'),
    ('2017-09-21'),
    ('2017-09-22')
)
select daterange, coalesce(sum(count), 0) as total_items
from (
    select daterange(lag(until_date) over (order by until_date), until_date)
    from ranges
    ) s
left join counts on date <@ daterange
where not lower_inf(daterange)
group by 1
order by 1;

        daterange        | total_items 
-------------------------+-------------
 [2017-09-08,2017-09-11) |           0
 [2017-09-11,2017-09-13) |           0
 [2017-09-13,2017-09-18) |           3
 [2017-09-18,2017-09-21) |          17
 [2017-09-21,2017-09-22) |           3
(5 rows)

注意，在上面的日期范围中，下限是包含的，而上限是独占的。

如果您想在日期范围内计算每日的项目：

select 
    daterange, total_items, 
    round(total_items::dec/(upper(daterange)- lower(daterange)), 2) as items_per_day
from (
    select daterange, coalesce(sum(count), 0) as total_items
    from (
        select daterange(lag(until_date) over (order by until_date), until_date)
        from ranges
        ) s
    left join counts on date <@ daterange
    where not lower_inf(daterange)
    group by 1
    ) s
order by 1

        daterange        | total_items | items_per_day 
-------------------------+-------------+---------------
 [2017-09-08,2017-09-11) |           0 |          0.00
 [2017-09-11,2017-09-13) |           0 |          0.00
 [2017-09-13,2017-09-18) |           3 |          0.60
 [2017-09-18,2017-09-21) |          17 |          5.67
 [2017-09-21,2017-09-22) |           3 |          3.00
(5 rows)

如何使用PostgreSQL

1 个答案: