每周查询和总计

Question

user, timeon, timeoff, hours 
Monday - 
1510, 2021-10-25 05:00:00, 2021-10-25 09:00:00 - 
1510, 2021-10-25 14:00:00, 2021-10-25 17:00:00 -

总数应该是 4 + 3 = 7

Tuesday - 
1510, 2021-10-25 05:00:00, 2021-10-25 09:00:00 - 
1510, 2021-10-25 14:00:00, 2021-10-25 17:00:00 -

总数应为 4 + 3 = 7，然后在本周末总数

我每天都有一个用户、timeon、timeoff，因为它们可能不止一次打开和关闭。我的总数在底部datediff(minute, timeon, timeoff)。我正在使用 Firebird 3.

我有一个虚拟机table

with M as
(
    select user, timeon, timeoff, hours
    from (select userid, timeon, timeoff,
    datediff(minute,timeon, timeoff)/60.00 as hours
    from sessions s
    where extract(weekday from timeon)=1
    and userID=1
    and cast(logon as date)='2021-10-28')
)
select * from M
union all
select '','','',sum(hours) from M

现在，我需要重复上面的内容，但星期二，然后是星期三，依此类推。我不能从另一个 v table 或联合开始。我做错了什么？

Answer 1

正如请求评论中提到的，我们通常会使用ROLLUP或GROUPING SETS，但Firebird不支持这些。

改用 WITH 子句是个好主意。这是包含小计和最终总计的查询：

with s as
(
  select
    extract(weekday from timeon) as day,
    userid, timeon, timeoff,
    datediff(minute, timeon, timeoff) / 60.00 as hours
  from sessions
  -- where userid = 1
)
select * from s
union all
select day, userid, null, null, sum(hours) from s group by day, userid
union all
select day, null, null, null, sum(hours) from s group by day
union all
select null, null, null , null, sum(hours) from s
order by day nulls last, userid nulls last, timeon nulls last;

这并不准确，因为我们将从晚上 10 点到凌晨 2 点的会话计为开始日的四个小时，而不是开始日的 2 小时和第二天的 2 小时。但是你在查询中做了同样的事情，所以我想这对你来说没问题。

每周查询和总计

Query and total for each week

sql

firebird

firebird-3.0