Question

我在mysql中有一个表格，其中列有日期，收入，以及当月每天产生的收入我需要把它变成格式日期，“此日期之前的收入总和”

我可以用

来做

select max(date) as date, sum(revenue) as total_revenue_to_date from table where dayofmonth(date)<=1
union
select max(date) as date, sum(revenue) as total_revenue_to_date from table where dayofmonth(date)<=2
.......

等，但想以更好的格式编写。

有人有什么想法吗？

asnwered：最短，最容易遵循：

SELECT fulldate, 
(SELECT SUM(margin) FROM fact_agg_margin_live_daily d2 WHERE d1.fulldate>=d2.fulldate) AS margin 
FROM fact_agg_margin_live_daily d1

Answer 1

我做了一些测试，所以这是我的例子：

表名= materiales 字段名称= id_material

行：1,2,3,4,6,7,9（列id_material）

使用的查询：

SELECT id_material，（SELECT SUM（id_material）FROM materiales M2 WHERE M1.id_material＆gt; = M2.id_material）AS suma 从materiales M1

预期结果：column1具有当前id，column2具有当前id的结果加上前一个的结果，是的，它可以工作。

您可以将其更改为当前场景，如下所示：

选择日期，（SELECT SUM（收入）FROM表T2 WHERE T1.date＆gt; = T2.date）AS revenue_until_current_date 从表T1

这将返回每个日期的收入，直到该日期为止。如果每天存储行，它将每天返回当天的收入加上前一天的收入。

Answer 2

将group by用于查询会更加简单：

select max(date) as date, sum(revenue) as total_revenue_to_date
from table
group by dayofmonth(date)

您想要累积总和（您的查询不会这样做）。这是一种方法（用30天的月份说明）：

select max(date) as date, sum(revenue) as total_revenue_to_date
from t join
     (select 0 as n union all select 1 union all select 2 union all select 3 union all select 4 union all
      select 5 as n union all select 6 union all select 7 union all select 8 union all select 9 union all
      select 10 as n union all select 11 union all select 12 union all select 13 union all select 14 union all
      select 15 as n union all select 16 union all select 17 union all select 18 union all select 19 union all
      select 20 as n union all select 21 union all select 22 union all select 23 union all select 24 union all
      select 25 as n union all select 26 union all select 27 union all select 28 union all select 29 union all
      select 30
     ) n
     on day(date) + n <= 30
group by day(date) + n;

您也可以使用变量执行此操作：

  select max(date) as date, sum(revenue) as total_revenue_to_date,
         @sum := @sum + SUM(revenue) as cumsum
  from table cross join
       (select @sum := 0) const
  group by dayofmonth(date)

每行的行日期之前的句点的Mysql总和

2 个答案: