SQL-使用频率（模式）计算激励

Question

我希望通过SQL查询来计算incentive列，并基于一个月内给学生补习的次数。

如果该学生每月在AP Math中接受过两次补习，则导师将获得20美元，否则为0美元。

我不想将日期分组为月份摘要，因此，我想按原样保留日期，因此我为每条记录每月分配两次辅导的学生分配10美元。

Answer 1

初始化数据

declare @StudentTable as table
(
    StudentVisitDate  date,
    StudentId int,
    StudentName varchar(100),
    [Subject] varchar (30),
    Tutor varchar(100),
    Incentive int
)

insert into @StudentTable
values
('2018-August-03',123,'Terry Brooks','AP Math','Shawn Green',10)
,('2018-August-04',123,'Terry Brooks','AP Science','Ted Berry',10)
,('2018-August-07',123,'Terry Brooks','Music','Shawn Green',10)
,('2018-September-03',123,'Terry Brooks','AP Math','Shawn Green',10)
,('2018-September-04',123,'Terry Brooks','AP Science','Ted Berry',10)
,('2018-September-07',123,'Terry Brooks','AP Math','Shawn Green',10)

获取每月访问次数，将数据分组

;with temp as
(
select
StudentId,  Subject,    Tutor,Month(StudentVisitDate) [month]
,max(StudentVisitDate) maxdate -- the date on which the incentive will be calculated
,Count(StudentVisitDate) [count]
from @StudentTable
Group by
StudentId,  Subject,    Tutor,Month(StudentVisitDate)
)

使用上表中的信息获取每个月的有效奖励措施

select
s.StudentVisitDate  ,
s.StudentId ,
s.StudentName ,
s.[Subject] ,
s.Tutor ,
Case
when t.maxdate = s.StudentVisitDate -- the incentive will be applied on the maximum date
then
 s.Incentive*t.count
 else
 0
 end Incentive
from @StudentTable s
inner join temp t
on s.StudentId=t.StudentId
and s.Subject =t.Subject
and s.Tutor=t.Tutor
and Month(s.StudentVisitDate)=t.month
order by StudentVisitDate

最终输出-

StudentVisitDate    StudentId   StudentName Subject Tutor   Incentive
2018-08-03  123 Terry Brooks    AP Math Shawn Green 10
2018-08-04  123 Terry Brooks    AP Science  Ted Berry   10
2018-08-07  123 Terry Brooks    Music   Shawn Green 10
2018-09-03  123 Terry Brooks    AP Math Shawn Green 0
2018-09-04  123 Terry Brooks    AP Science  Ted Berry   10
2018-09-07  123 Terry Brooks    AP Math Shawn Green 20

Answer 2

我认为您只需要窗口函数：

select t.*,
       (case when cnt = 2 then 10 else 0 end) as incentive
from (select t.*,
             count(*) over (partition by studentid, subject, tutor, year(studentvisitdate), month(studentvisitdate) as cnt
      from t
     ) t