按周分组的记录数的累积总和

Question

我有以下数据库架构

ID  creation_date
1      2019-06-03
2      2019-06-04
3      2019-06-04
4      2019-06-10
5      2019-06-11

我需要按周找出表组的总大小。我正在寻找的输出类似于

year   week number_of_records
2019    23      3
2019    24      5

我正在编写以下查询，它只给我每周创建的记录数

> select year(creation_date) as year, weekofyear(creation_date) as week,
> count(id) from input group by year, week;

我得到的输出是

year   week number_of_records
2019    23      3
2019    24      2

Answer 1

你似乎想要一个累积的总和。您可以直接在聚合查询中使用窗口函数执行此操作：

select year(creation_date) as year, weekofyear(creation_date) as week,
       count(*) as number_of_starts,
       sum(count(*)) over (order by min(creation_date)) as number_of_records
from input
group by year, week;

Answer 2

看看窗口（或分析）函数。 与聚合函数不同，窗口函数保留结果行并促进与它们相关的操作。在over子句中使用order by时，按照指定的顺序从第一行到当前行开窗，正是你需要的。

select year, week, sum(number_of_records) over (order by year, week)
from (
  select year(creation_date) as year, weekofyear(creation_date) as week,
  count(id) as number_of_records
  from input group by year, week
) your_sql

我猜你还需要重置每年的总和，我留给你做练习（提示：partition 子句）。

Answer 3

对于 8.0 之前的版本...

架构 (MySQL v5.7)

CREATE TABLE my_table
(ID SERIAL PRIMARY KEY
,creation_date DATE NOT NULL
);

INSERT INTO my_table VALUES
(1    ,  '2019-06-03'),
(2     , '2019-06-04'),
(3     , '2019-06-04'),
(4      ,'2019-06-10'),
(5      ,'2019-06-11');

---

查询 #1

SELECT a.yearweek
, @i:=@i+a.total running
FROM
(SELECT DATE_FORMAT(x.creation_date,'%x-%v') yearweek
, COUNT(*) total
FROM my_table x
GROUP BY yearweek
)a
JOIN (SELECT @i:=0) vars
ORDER BY a.yearweek;

| yearweek | running |
| -------- | ------- |
| 2019-23  | 3       |
| 2019-24  | 5       |

---

View on DB Fiddle