在 mysql 中使用 count(*) .. Over(*)答案

【问题标题】：Using count(*) .. Over(*) in mysql在 mysql 中使用 count(*) .. Over(*)
【发布时间】：2020-12-08 14:39:52
【问题描述】：

我的数据如下所示，

 requestedDate         Status
2020-04-21            APPROVED
2020-04-23            APPROVED
2020-04-27            PENDING
2020-05-21            PENDING
2020-06-01            APPROVED

我想提取如下所示的报告，其中按状态和月份进行计数。

Status      StatusCount Month   MonthCount  CountTotal
APPROVED        2        APR        3          5
PENDING         1        MAY        1          5
APPROVED        1        JUN        1          5

我的 sql 如下所示，

    select distinct
    status,
    count(status) over (partition by status) as total_by_status,
    CASE
        WHEN Month(requestedDate) = 1 THEN 'JAN'
        WHEN Month(requestedDate) = 2 THEN 'FEB'
        WHEN Month(requestedDate) = 3 THEN 'MAR'
        WHEN Month(requestedDate) = 4 THEN 'APR'
        WHEN Month(requestedDate) = 5 THEN 'MAY'
        WHEN Month(requestedDate) = 6 THEN 'JUN'
        WHEN Month(requestedDate) = 7 THEN 'JUL'
        WHEN Month(requestedDate) = 8 THEN 'AUG'
        WHEN Month(requestedDate) = 9 THEN 'SEP'
        WHEN Month(requestedDate) = 10 THEN 'OCT'
        WHEN Month(requestedDate) = 11 THEN 'NOV'
        WHEN Month(requestedDate) = 12 THEN 'DEC'
    END AS myMONTH,
  count(Month(requestedDate)) over (partition by Month(requestedDate)) as total_by_month,
  count(*) over () as Totals
from Reports
where
 requestedDate between DATE_SUB(CURDATE(), INTERVAL 120 DAY)  and date(CURDATE())
order by 1;

输出看起来像，

status    total_by_status   myMONTH total_by_month  Totals
APPROVED    3                APR        3           5
APPROVED    3                JUN        1           5
PENDING     2                APR        3           5
PENDING     2                MAY        1           5

dbfiddle

【问题讨论】：

标签： mysql sql datetime count window-functions

【解决方案1】：

首先，您需要一个有效的聚合查询。然后你可以在它上面使用窗口函数（在这里，你通常会计算窗口的计数总和）。

我会这样写：

select 
    status,
    count(*) status_count,
    date_format(requestedDate, '%b') requested_month
    sum(count(*)) over(partition by year(requestedDate), month(requestedDate)) month_count,
    sum(count(*)) over() total_count
from reports
where requestedDate between current_date - interval 120 day and current_date
group by status, year(requestedDate), month(requestedDate), date_format(requestedDate, '%b')

【讨论】：

我在 requested_month 之后添加了一个 ',' 但效果很好，谢谢

【解决方案2】：

由于它只是过去 120 天（不会发生去年同月），所以我们也可以使用 distinct 而不是 group by），如下所示：

    select distinct status, 
           count(*) over (partition by status) as total_by_status,
           date_format(requestedDate, '%b') mymonth,
           count(Month(requestedDate)) over (partition by Month(requestedDate)) as total_by_month,
           count(*) over () as total_by_month
    from reports
    where requestedDate between current_date - interval 120 day and current_date
order by status, mymonth

Demo

【讨论】：