我想根据表格日期的工作日平均取消特定查询的值,并在其他日子中复制。我知道这种解释令人困惑,请参见下面的示例。
Eg:- Table Structure -
ID, Value, Date
Expected Result-
Avg(value), Weekname(day)
12.23 , Mon
12.23 , Tue
12.23 , Wed
12.23 , Thu
12.23 , Fri
34.56 , Sat
56.34 , Sun
ie,周一至周五的价值是相同的。
Current Result
Avg(value), Weekname(day)
12.23 , Mon
34.56 , Sat
56.34 , Sun
查询看起来像这样 -
select avg(value)
,daynameofweek(date)
from table
group by (CASE WHEN DAYOFWEEK(DATE) NOT BETWEEN 2 and 6 THEN DAYOFWEEK(DATE) END)
感谢您的帮助。
编辑 - 添加示例数据集
ID Value Date
1 2.500 2017-01-01
2 0.674 2017-01-02
3 2.743 2017-01-03
4 1.460 2017-01-04
5 1.457 2017-01-05
6 1.791 2017-01-06
7 1.896 2017-01-07
8 2.015 2017-01-08
9 2.224 2017-01-09
10 1.635 2017-01-10
11 1.100 2017-01-11
12 0.441 2017-01-12
13 0.809 2017-01-13
14 1.508 2017-01-14
预期结果
Avg(Value) daynameofweek(date)
1.625 Monday
1.625 Tuesday
1.625 Wednesday
1.625 Thursday
1.625 Friday
1.300 Saturday
1.702 Sunday
您表明您最终在周一至周五的'Mon'中最终出现了,但这是由DBMS任意选择的,您也可以最终以'Wed'的方式选择,因为这一天也在分组范围内。因此,首先在此处获得确定值。然后加入虚构的日子表:
select
days.dayname,
data.avg_value
from
(
select
case when dayofweek(date) between 2 and 6 then 2 else dayofweek(date) end as day
avg(value) as avg_value
from table
group by case when dayofweek(date) between 2 and 6 then 2 else dayofweek(date) end
) data
join
(
select 1 as day, 'SUN' as dayname, 7 as sortkey union all
select 2 as day, 'MON' as dayname, 1 as sortkey union all
select 2 as day, 'TUE' as dayname, 2 as sortkey union all
select 2 as day, 'WED' as dayname, 3 as sortkey union all
select 2 as day, 'THU' as dayname, 4 as sortkey union all
select 2 as day, 'FRI' as dayname, 5 as sortkey union all
select 7 as day, 'SAT' as dayname, 6 as sortkey
) days on days.day = data.day
order by data.sortkey;
(如果我没记错的话,MySQL允许按别名名称(group by day
)进行分组,这不是标准的。