我有一个
Android应用程序.在我的Android应用程序中,我有一个sqlite数据库.在我的sqlite数据库中,我有一个如下所示的表：
_id a b c d 1 1 1 1 0 2 0 1 1 1 3 1 0 0 1 4 0 1 0 1
我想为每一行计算：在当前或上一行中有一个1的所有列a,b,c和d,当前和上一行中有一个1的百分比是多少？输出看起来像这样：
_id a b c d result 1 1 1 1 0 NULL 2 0 1 1 1 50% 3 1 0 0 1 25% 4 0 1 0 1 33%
解决方法
Select CurAndNext.T1_id,Sum( Case When D1.Val + D2.Val = 1 Then 1 End ) As CntInEitherRow,Sum( Case When D1.Val + D2.Val = 2 Then 1 End ) / 4.000 As PercBoth
From (
Select T1._id As T1_id,Max( T2._id ) As T2_id
From MyTable As T1
Left Join MyTable As T2
On T2._id < T1._id
Group By T1._id
) As CurAndNext
Join (
Select _id,'a' As Col,a As Val From MyTable As T1
Union All
Select _id,'b',b From MyTable As T1
Union All
Select _id,'c',c From MyTable As T1
Union All
Select _id,'d',d From MyTable As T1
) As D1
On D1._id = CurAndNext.T1_id
Left Join (
Select _id,a As Val From MyTable As T1
Union All
Select _id,b From MyTable As T1
Union All
Select _id,c From MyTable As T1
Union All
Select _id,d From MyTable As T1
) As D2
On D2._id = CurAndNext.T2_id
And D2.Col = D1.Col
Group By CurAndNext.T1_Id
使查询困难的一个重要因素是数据被非规范化.因此,我必须将其标准化才能获得您所寻求的信息.
知道a,c和d列代表哪些列在世界上有所不同.上述查询的复杂性表明模式不能很好地映射到业务需求.知道他们代表学生出勤,我们可以设计一个替代模式.
Create Table Student
(
Id int not null Primary Key,Name varchar(50) not null
)
Create Table Class
(
Id int not null Primary Key,Name varchar(50) not null
)
-- if using dates,this would be the equivalent
-- of a calendar table
Create Table ClassDay
(
DayNum int not null Primary Key
)
-- ClassDayNum would be better as a Date
Create Table Attendence
(
StudentId int References Student( Id ),ClassId int References Class( Id ),ClassDayNum int not null References ClassDay( DayNum ),Unique( StudentId,ClassId,ClassDayNum )
)
Insert Student( Id,Name )
Select 1,'a'
Union All Select 2,'b'
Union All Select 3,'c'
Union All Select 4,'d'
Insert Class( Id,Name )
Values (1,'Some Class' )
Insert ClassDay( DayNum )
Select 1
Union All Select 2
Union All Select 3
Union All Select 4
Insert Attendence( ClassId,StudentId,ClassDay )
Select 1,1,1
Union All Select 1,3
Union All Select 1,2,2
Union All Select 1,4
Union All Select 1,3,4,4
of all the columns a,c and d that have a 1 in EITHER the current or prevIoUs row
您的结果实际读取的方式实际上是请求在一天而不是之前或前一天参加的人数而不是当前人数.
Select Class.Id,ClassDay.DayNum,Count(distinct A.StudentId) As Attendence,Count(distinct A.StudentId) / 4.000 As Ratio From Class Cross Join Student Cross Join ClassDay Left Join Attendence As A On A.ClassId = Class.Id And A.StudentId = Student.Id And A.ClassDayNum = ClassDay.DayNum And A.ClassDayNum > 1 Left Join Attendence As A2 On A2.ClassId = Class.Id And A2.StudentId = Student.Id And A2.ClassDayNum = ClassDay.DayNum - 1 Where Not( A.StudentId Is Not Null And A2.StudentId Is Not Null ) Group By Class.Id,ClassDay.DayNum结果：
DayNum Attendence | Ratio 1 | 0 | 0 2 | 1 | .25 3 | 1 | .25 4 | 1 | .25what percentage have a 1 in BOTH the current and prevIoUs row
Select ClassDay.DayNum,Sum( Case When A.StudentId Is Not Null And A2.StudentId Is Not Null Then 1 End ),Sum( Case When A.StudentId Is Not Null And A2.StudentId Is Not Null Then 1 End ) / 4.000 From Class Cross Join Student Cross Join ClassDay Left Join Attendence As A On A.ClassId = Class.Id And A.StudentId = Student.Id And A.ClassDayNum = ClassDay.DayNum And A.ClassDayNum > 1 Left Join Attendence As A2 On A2.ClassId = Class.Id And A2.StudentId = Student.Id And A2.ClassDayNum = ClassDay.DayNum - 1 Group By ClassDay.DayNumDayNum | Attendence | Ratio 1 | NULL | NULL 2 | 2 | 0.500000 3 | 1 | 0.250000 4 | 1 | 0.250000
