SQL Server 条件比较
我有两个表:
CREATE TABLE #HOURS
(DAY INTEGER,
HOUR INTEGER)
CREATE TABLE #PERSONS
(DAY INTEGER, HOUR INTEGER,
Name NVARCHAR(50))
GO
INSERT #HOURS VALUES (1, 5)
INSERT #HOURS VALUES (1, 6)
INSERT #HOURS VALUES (1, 8)
INSERT #HOURS VALUES (1, 10)
INSERT #HOURS VALUES (1, 14)
INSERT #HOURS VALUES (1, 15)
INSERT #HOURS VALUES (1, 16)
INSERT #HOURS VALUES (1, 17)
INSERT #HOURS VALUES (1, 18)
INSERT #PERSONS VALUES (1, 5, 'Steve')
INSERT #PERSONS VALUES (1, 6, 'Steve')
INSERT #PERSONS VALUES (1, 7, 'Steve')
INSERT #PERSONS VALUES (1, 8, 'Steve')
INSERT #PERSONS VALUES (1, 10, 'Steve')
INSERT #PERSONS VALUES (1, 14, 'Steve')
INSERT #PERSONS VALUES (1, 15, 'Steve')
INSERT #PERSONS VALUES (1, 16, 'Steve')
INSERT #PERSONS VALUES (1, 17, 'Steve')
INSERT #PERSONS VALUES (1, 10, 'Jim')
INSERT #PERSONS VALUES (1, 11, 'Jim')
INSERT #PERSONS VALUES (1, 12, 'Jim')
INSERT #PERSONS VALUES (1, 13, 'Jim')
GO
Hours 显示工作时间,#Persons 显示按小时进入系统的人员。 我想找到工作时间与工作时间表相符的人。但他或她可以跳过两个工作时间。
我已经尝试过:
select t.Day, sum(t.Nulls)
from
(select h.Hour, h.Day
, Case
WHEN p.Hour is null Then 1 ELSE 0 END Nulls
from #HOURS h
left join #PERSONS P on h.Hour = p.Hour AND h.Day = p.Day) t
group by t.Day
HAVING sum(t.Nulls) < 2
但这仅在同一天没有不同的人时才有效;)
有什么建议吗?
I have two tables:
CREATE TABLE #HOURS
(DAY INTEGER,
HOUR INTEGER)
CREATE TABLE #PERSONS
(DAY INTEGER, HOUR INTEGER,
Name NVARCHAR(50))
GO
INSERT #HOURS VALUES (1, 5)
INSERT #HOURS VALUES (1, 6)
INSERT #HOURS VALUES (1, 8)
INSERT #HOURS VALUES (1, 10)
INSERT #HOURS VALUES (1, 14)
INSERT #HOURS VALUES (1, 15)
INSERT #HOURS VALUES (1, 16)
INSERT #HOURS VALUES (1, 17)
INSERT #HOURS VALUES (1, 18)
INSERT #PERSONS VALUES (1, 5, 'Steve')
INSERT #PERSONS VALUES (1, 6, 'Steve')
INSERT #PERSONS VALUES (1, 7, 'Steve')
INSERT #PERSONS VALUES (1, 8, 'Steve')
INSERT #PERSONS VALUES (1, 10, 'Steve')
INSERT #PERSONS VALUES (1, 14, 'Steve')
INSERT #PERSONS VALUES (1, 15, 'Steve')
INSERT #PERSONS VALUES (1, 16, 'Steve')
INSERT #PERSONS VALUES (1, 17, 'Steve')
INSERT #PERSONS VALUES (1, 10, 'Jim')
INSERT #PERSONS VALUES (1, 11, 'Jim')
INSERT #PERSONS VALUES (1, 12, 'Jim')
INSERT #PERSONS VALUES (1, 13, 'Jim')
GO
Hours shows the work hours and #Persons shows the persons that entered the system on hourly base.
I'd like to find the persons whose work hours matches the hours table. But he or she can skip two work hours.
I've tried this:
select t.Day, sum(t.Nulls)
from
(select h.Hour, h.Day
, Case
WHEN p.Hour is null Then 1 ELSE 0 END Nulls
from #HOURS h
left join #PERSONS P on h.Hour = p.Hour AND h.Day = p.Day) t
group by t.Day
HAVING sum(t.Nulls) < 2
But this only works when there is not different persons on the same day ;)
Any suggestions?
如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱,如果其他用户或者作者回复了您的评论,将不能在第一时间通知您!
发布评论
评论(2)
我认为您的意思是,对于日期和人员的每个组合,如果该人当天工作的有效小时数在有效小时数的 2 小时之内,您希望返回该值那天存在。如果是这样,这应该可以解决问题:
What I think you mean is that for each combination of day and person, you want to return it if the number of valid hours that that person worked on that day is within 2 hours of the number of valid hours that exist for that day. If so, this should do the trick:
我认为您仍然需要进一步澄清您期望的结果。与此同时,这会让你开始吗?
I think you still need to clarify a bit further what you'd expect as a result. In the mean time, would this get you started?