Mysql 连续日期的计数总和
我需要得到每个月一个项目的总数 到目前为止,使用以下代码:Mysql 连续日期的计数总和,mysql,date,count,sum,Mysql,Date,Count,Sum,我需要得到每个月一个项目的总数 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 Mont
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
我希望得到的不是每个月的数字,而是每个月开始时的总和,大致如下:
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,因此子请求几乎是一个禁忌:x您可以首先选择不同的year.month,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
如果您想每年重置,只需将比较更改为“年与年之间相等”,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
您可以先选择不同的年.月,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
如果您想每年重置,只需将比较更改为“年与年之间相等”,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
使用相关子查询计算日期早于当前查询的行数也应该有效:
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);
在操作中显示它使用相关子查询计算日期早于当前日期的行数也应该有效:
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);
在行动中展示它类似的东西可能会奏效:
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);
诀窍是自动加入前几个月的所有数据,并从加入的表中计数。类似的方法可能会奏效:
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);
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)
诀窍是自动加入前几个月的所有数据,并从已加入的表中进行计数
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)
这应该对你有好处
这应该能帮你,这是一个很好的开始,谢谢你!然而,结果是2014年的总和,然后在2015年重置,这意味着我得到了20145->8->20153->10->16。。。不是20145->8->201511->18->24…我已经为您更新了我的答案,带有重置查询。这是一个非常好的开始,谢谢您!然而,结果是2014年的总和,然后在2015年重置,这意味着我得到了20145->8->20153->10->16。。。不是20145->8->201511->18->24…我已经为您更新了我的答案,重置查询我在计数中得到了非常愚蠢的结果,但是我喜欢这个解决方案,因为我需要将其实现为一个视图,所以子请求并不是一件好事:pAlright,给我一秒钟,我将创建一些测试数据。该查询有点失控,但我在插入一些测试数据后编辑了我的答案。顺便说一句,如果您担心性能,那么加入函数是不可能的,因为MySQL将无法使用索引。感谢您的时间,我将在周一与您联系。我所有的东西都在工作,我刚刚回家过了一个轻松的周末^^我在计数中得到了非常愚蠢的结果,但是我喜欢这个解决方案,因为我需要将其作为一个视图来实现,所以子请求对于这一点来说并不是一件好事:好吧,给我一点时间,我将创建一些测试数据。那个查询有点失控,但是我在插入了一些测试数据后编辑了我的答案。顺便说一句,如果你担心性能,那么加入函数是不行的,因为MySQL将无法使用索引。谢谢你的时间,我会在周一回复你。我工作时什么都有,我刚回家过了一个轻松的周末