复制MySQL查询的结果集

我希望根据表中日期的工作日平均出特定查询的值，并在其他日期复制它。我知道这个解释令人困惑，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

也就是说，周一到周五的值是相同的

Current Result
Avg(value), Weekname(day)
12.23    , Mon
34.56    , Sat
56.34    , Sun

查询如下所示-

select avg(value)
      ,daynameofweek(date) 
from table
group by (CASE WHEN  DAYOFWEEK(DATE) NOT BETWEEN 2 and 6 THEN DAYOFWEEK(DATE) END)

提前谢谢你的帮助

编辑-添加示例数据集 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

---

您显示的结果是周一到周五的“MON”，但这是DBMS任意选择的，您也可以以“WED”结束，因为这一天也在分组范围内。所以从这里得到一个确定的值开始。然后加入一个组成的日表：

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允许按别名分组（

按天分组

），这不符合标准。如果您更喜欢，可以使用它。）

请在postWhy 2和6中包含示例数据集和表定义？这背后的逻辑是什么？复制值直到间隙结束？请重新表述问题。2指星期一的工作日，6指星期五。请参阅：示例数据集似乎与原始问题并不相关

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;