MS SQL - 老化记录

Question

请帮助，因为我坚持这一点！ ：/

我有5列，我在查询选择和最后2导出/计算

Date | Account | Symbol | Type | User | AgeKEY | Age |  

，其中关键是级联（帐号+符号+型号+用户）

我如何回顾历史1年并计算记录的年龄？年龄是工作日连续＃该AgeKey出现在历史

老化逻辑实施例 -

11/3 KeyExists hence Age = 1 

11/4 KeyExists hence Age = 2 

11/7 KeyExists hence Age = 3 (note over weekend ages only by 1 day) 

11/8 KeyDoesntExist 

11/9 KeyExists hence Age = 1 (counter restarts from 1 if this happens) 

Answer 1

与T-SQL环路（它从标签表中读取数据，并插入到tab_result）：

create table tab 
(dt date, id int); 

insert into tab values(DATEADD(day,-12,GETDATE()),1); 
insert into tab values(DATEADD(day,-10,GETDATE()),1); 
insert into tab values(DATEADD(day,-9,GETDATE()),1); 
insert into tab values(DATEADD(day,-8,GETDATE()),1); 
insert into tab values(DATEADD(day,-7,GETDATE()),3); 
insert into tab values(DATEADD(day,-6,GETDATE()),3); 
insert into tab values(DATEADD(day,-5,GETDATE()),1); 
insert into tab values(DATEADD(day,-4,GETDATE()),1); 
insert into tab values(DATEADD(day,-3,GETDATE()),1); 

create table tab_result 
(dt date, id int, age int); 

DECLARE @id INT, @dt date, @prevId INT=NULL, @prevDt date=NULL, @age int 

DECLARE CurName CURSOR FAST_FORWARD READ_ONLY 
FOR 
    SELECT id,dt 
    FROM tab 
    ORDER BY id,dt 
OPEN CurName 


FETCH NEXT FROM CurName INTO @id, @dt 
set @age=0; 
WHILE @@FETCH_STATUS = 0 
    BEGIN 
     if (@prevId<>@id or @prevDt <> DATEADD(day,-1, @dt)) 
       set @age=1; 
      else 
       set @age=@age+1; 
     insert into tab_result values (@dt, @id, @age) 

     set @prevId=@id 
     set @prevDt=@dt 
     FETCH NEXT FROM CurName INTO @id, @dt 

    END 

CLOSE CurName 
DEALLOCATE CurName 

select * from tab_result order by id, dt; 

与普通的SQL它会像下面（ID在本例中是你的钥匙）：

create table tab 
(dt date, id int); 

insert into tab values(DATEADD(day,-12,GETDATE()),1); 
insert into tab values(DATEADD(day,-10,GETDATE()),1); 
insert into tab values(DATEADD(day,-9,GETDATE()),1); 
insert into tab values(DATEADD(day,-8,GETDATE()),1); 
insert into tab values(DATEADD(day,-7,GETDATE()),3); 
insert into tab values(DATEADD(day,-6,GETDATE()),3); 
insert into tab values(DATEADD(day,-5,GETDATE()),1); 
insert into tab values(DATEADD(day,-4,GETDATE()),1); 
insert into tab values(DATEADD(day,-3,GETDATE()),1); 

with x as ( 
select tab.dt, 
    tab.id, 
    case when prev.dt is not null then 1 else 0 end as exists_on_prev_day 
    from 
    tab left outer join tab prev on (tab.id=prev.id and DATEADD(day,-1 , tab.dt)= prev.dt) 
) 
select id,dt, 
(select 
    -- count all records with the same id and date less or equal date of the given record 
count(*) from x x2 where x2.id=x.id and x2.dt<=x.dt 
-- (tricky part) we want to count only records between current record and last record without "previous" record (that is with exists_on_prev_day flag = 0) 
    and not exists (select 1 from x x3 where x3.id=x2.id and x3.dt>x2.dt and x3.dt<=x.dt and x3.exists_on_prev_day=0)) age 
from x 
order by id, dt; 

结果：

id dt     age 
1 1 19.11.2016 00:00:00 1 
2 1 21.11.2016 00:00:00 1 
3 1 22.11.2016 00:00:00 2 
4 1 23.11.2016 00:00:00 3 
5 1 26.11.2016 00:00:00 1 
6 1 27.11.2016 00:00:00 2 
7 1 28.11.2016 00:00:00 3 
8 3 24.11.2016 00:00:00 1 
9 3 25.11.2016 00:00:00 2