mysql每周分组，每周添加总计

我正在尝试创建一个查询，该查询将返回基于每周的数字总数，以便创建一个上升趋势折线图。在我的表中，有许多记录记录了完成日期（已完成）。我希望能够创建一个查询，每周生成一个滚动总数。因此，如果第1周完成了10个，第2周完成了15个，第3周完成了5个，那么期望的结果将是：

SET @runtot:=0;
SELECT
   DATE_FORMAT(completed,'%d/%m/%Y') AS niceDate,
   wk,
   (@runtot := @runtot + c) AS rt
FROM
   (SELECT
        completed,
   yearweek(completed)as wk,
   COUNT(*) AS c
FROM  `w10_upgrades`
where status = 'Successful' and type = 'Normal'
and yearweek(completed) is not null
GROUP  BY wk
ORDER  BY wk) x

第一周总计：10 第2周总计：25 第3周总计：30

SET @runtot:=0;
SELECT
   DATE_FORMAT(completed,'%d/%m/%Y') AS niceDate,
   wk,
   (@runtot := @runtot + c) AS rt
FROM
   (SELECT
        completed,
   yearweek(completed)as wk,
   COUNT(*) AS c
FROM  `w10_upgrades`
where status = 'Successful' and type = 'Normal'
and yearweek(completed) is not null
GROUP  BY wk
ORDER  BY wk) x

样本数据：

 id status  sched
 12 Successful  2017-04-04 00:00:00.000
 15 Successful  2017-06-20 19:30:00.000
 18 Successful  2017-10-17 18:00:00.000
 26 Successful  2017-04-05 00:00:00.000
 29 Successful  2017-06-16 00:00:00.000
 30 Successful  2017-04-06 00:00:00.000
 31 Successful  2017-04-07 00:00:00.000
 32 Successful  2017-04-06 00:00:00.000
 34 Successful  2017-10-18 18:00:00.000
 35 Successful  2017-06-13 00:00:00.000

SET @runtot:=0;
SELECT
   DATE_FORMAT(completed,'%d/%m/%Y') AS niceDate,
   wk,
   (@runtot := @runtot + c) AS rt
FROM
   (SELECT
        completed,
   yearweek(completed)as wk,
   COUNT(*) AS c
FROM  `w10_upgrades`
where status = 'Successful' and type = 'Normal'
and yearweek(completed) is not null
GROUP  BY wk
ORDER  BY wk) x

这是我用来在没有任何汇总的情况下按周成功生成数据的查询。我试着加上“WITH ROLLUP”，但最后只给出了总计，而不是每周

select DATE_FORMAT(completed,'%d/%m/%Y') AS nd , wk, count(*)as totals
from
(
   select id, completed, yearweek(completed)as wk from w10_upgrades
   where status = 'Successful' and type = 'Normal'
   and yearweek(completed) is not null
) x
GROUP BY wk
ORDER BY wk;

SET @runtot:=0;
SELECT
   DATE_FORMAT(completed,'%d/%m/%Y') AS niceDate,
   wk,
   (@runtot := @runtot + c) AS rt
FROM
   (SELECT
        completed,
   yearweek(completed)as wk,
   COUNT(*) AS c
FROM  `w10_upgrades`
where status = 'Successful' and type = 'Normal'
and yearweek(completed) is not null
GROUP  BY wk
ORDER  BY wk) x

期望输出：

 wk     totals
 201714   10
 201715   25 (output would = week 201714 + 201715)
 201716   55 (output would = week 201714 + 201715 + 201716)
 ect...

SET @runtot:=0;
SELECT
   DATE_FORMAT(completed,'%d/%m/%Y') AS niceDate,
   wk,
   (@runtot := @runtot + c) AS rt
FROM
   (SELECT
        completed,
   yearweek(completed)as wk,
   COUNT(*) AS c
FROM  `w10_upgrades`
where status = 'Successful' and type = 'Normal'
and yearweek(completed) is not null
GROUP  BY wk
ORDER  BY wk) x

任何方向都值得赞赏。我找不到与此相关的任何内容。

最终结果

SET @runtot:=0;
SELECT
   DATE_FORMAT(completed,'%d/%m/%Y') AS niceDate,
   wk,
   (@runtot := @runtot + c) AS rt
FROM
   (SELECT
        completed,
   yearweek(completed)as wk,
   COUNT(*) AS c
FROM  `w10_upgrades`
where status = 'Successful' and type = 'Normal'
and yearweek(completed) is not null
GROUP  BY wk
ORDER  BY wk) x

---

请显示您的模式、示例源数据和预期输出。谢谢。像这样的东西可能会对你有帮助：看起来很有希望。谢谢