Question

我需要每个月获得一件商品的总数。

到目前为止，使用以下代码：

Select Count(mpay_collector_company.id) As `Number of Collector Companies`,
  Month(mpay_collector_company.created_at) As Month,
  Year(mpay_collector_company.created_at) As Year
From mpay_collector_company
Group By Month(mpay_collector_company.created_at),
  Year(mpay_collector_company.created_at)

我有以下回应：

#|Year|Month
------------
5|2014|11
3|2014|12
3|2015|1
7|2015|2
6|2015|3
2|2015|4
5|2015|6
1|2015|7

而不是每个月的数字，我想从每个月的开头得到总和，这将是这样的：

Sum|Year|Month
--------------
  5|2014|11
  8|2014|12
 11|2015|1
 18|2015|2
 24|2015|3
 26|2015|4
 31|2015|6
 32|2015|7

有什么想法吗？

编辑：此请求将作为一个视图实现，因此子请求几乎是不可行的：x

Answer 1

您可以先选择不同的year.months，然后检查每个值的计数

select year,month , 
    ( SELECT count(*) FROM mpay_collector_company WHERE 
        Year(mpay_collector_company.created_at) <= temp1.year 
        AND 
        Month(mpay_collector_company.created_at) <= temp1.month 
    )
from (
Select DISTINCT 
    Month(mpay_collector_company.created_at) As month,
    Year(mpay_collector_company.created_at) As year
From mpay_collector_company 
) as temp1
ORDER BY year,month

如果你想＆＃34;重置＆＃34;对于每一年，只需将年份之间的比较改为EQUAL即可

select year,month , 
    ( SELECT count(*) FROM mpay_collector_company WHERE 
        Year(mpay_collector_company.created_at) = temp1.year 
        AND 
        Month(mpay_collector_company.created_at) <= temp1.month 
    )
from (
Select DISTINCT 
    Month(mpay_collector_company.created_at) As month,
    Year(mpay_collector_company.created_at) As year
From mpay_collector_company 
) as temp1
ORDER BY year,month

Answer 2

使用相关子查询来计算当前应该也可以使用日期的行数：

Select 
  Count(id) As "Number of Collector Companies",
  Month(created_at) As Month,
  Year(created_at) As Year,
  (
    Select Count(*) 
    From mpay_collector_company 
    Where EXTRACT(YEAR_MONTH from created_at) <= EXTRACT(YEAR_MONTH from m.created_at)     
  ) as running_count
From mpay_collector_company m
Group By Year(created_at), Month(created_at);

Sample SQL Fiddle在行动中显示

Answer 3

这样的事情可行：

Select Count(distinct previous_mpay_collector_company.id)+ Count(distinct mpay_collector_company.id) As `Number of Collector Companies`,
       Month(mpay_collector_company.created_at) As Month,
       Year(mpay_collector_company.created_at) As Year
From mpay_collector_company
  Left Join mpay_collector_company previous_mpay_collector_company
    On   Year(mpay_collector_company.created_at) > Year(previous_mpay_collector_company.created_at)
 OR (Month(mpay_collector_company.created_at) > Month(previous_mpay_collector_company.created_at)
         And  Year(mpay_collector_company.created_at) >= Year(previous_mpay_collector_company.created_at))
  And previous_mpay_collector_company.id <> mpay_collector_company.id
Group By Month(mpay_collector_company.created_at),
Year(mpay_collector_company.created_at);

诀窍是自我加入前几个月并从联合表中算起。

Answer 4

    Select 
(Select Sum(Count(Z.id)) FROM mpay_collector_company z 
where  z.created_at <= A.created_at) as '#', 
  Month(A.created_at) As Month,
  Year(A.created_at) As Year
From mpay_collector_company A
Group By Month(mpay_collector_company.created_at),
  Year(mpay_collector_company.created_at)

这应该为你做，

连续日期的计数总和

4 个答案: