Sql 时间序列“；“地位”；计算

样本数据：

rep_signup_date rep_id client_registration_date client_id 
1/2/2018        1      1/5/2018                 1          
1/2/2018        1      1/9/2018                 2
1/2/2018        1      2/15/2018                3
1/4/2018        2      2/3/2018                 4
1/4/2018        2      3/9/2018                 5
2/1/2018        3      2/2/2018                 6

rep_signup_date rep_id month    status
1/2/2018        1      1/1/2018 2
1/2/2018        1      2/1/2018 3     
1/4/2018        2      2/1/2018 1
1/4/2018        2      3/1/2018 2
2/1/2018        3      2/1/2018 1

我们对代表“状态”进行分类的方式是基于客户数量： 1个客户机-状态1，2个客户机-状态2，3+客户机-状态3，因此目前我们了解以下情况：

select rep_signup_date, rep_id,  
case when count(client_id) over (partition by rep_id) >=3 then '3'
     when count(client_id)  over (partition by rep_id) =2 then '2'
     when count(client_id)  over (partition by rep_id) =1 then '1'
     end status
from reps r
left join clients c on c.rep_id=r.id

rep_signup_date rep_id  status
1/2/2018        1       3     
1/4/2018        2       2
2/1/2018        3       1

但是，这些状态截至当前日期；我尝试为月份添加

date\u trunc（'month'，client\u registration\u date）：：date

，但它仍然根据最大日期提供当前快照数据，而不是静态时间点

我希望能够在每个月底获得状态-例如，1月底的RepID1是状态2

预期输出：

rep_signup_date rep_id client_registration_date client_id 
1/2/2018        1      1/5/2018                 1          
1/2/2018        1      1/9/2018                 2
1/2/2018        1      2/15/2018                3
1/4/2018        2      2/3/2018                 4
1/4/2018        2      3/9/2018                 5
2/1/2018        3      2/2/2018                 6

rep_signup_date rep_id month    status
1/2/2018        1      1/1/2018 2
1/2/2018        1      2/1/2018 3     
1/4/2018        2      2/1/2018 1
1/4/2018        2      3/1/2018 2
2/1/2018        3      2/1/2018 1

---

我怎么去那里？谢谢。

使用

订购人

：

select rep_signup_date, rep_id,  
       (case when count(client_id) over (partition by rep_id order by client_registration_date rows between unbounded preceding and current row) >= 3 then '3'
             when count(client_id) over (partition by rep_id order by client_registration_date rows between unbounded preceding and current row) = 2 then '2'
             when count(client_id) over (partition by rep_id order by client_registration_date rows between unbounded preceding and current row) = 1 then '1'
        end) as status
from reps r left join
     clients c
     on c.rep_id = r.id;

每个客户机/代表似乎有一行，因此使用

行数（）

比使用累积计数要简单得多：

select rep_signup_date, rep_id,  
       (case when row_number() over (partition by rep_id order by client_registration_date ) >= 3 then '3'
             when row_number() over (partition by rep_id order by client_registration_date rows) = 2 then '2'
             when row_number() over (partition by rep_id order by client_registration_date = 1 then '1'
        end) as status
from reps r left join
     clients c
     on c.rep_id = r.id;

这可以进一步简化为：

select rep_signup_date, rep_id,  
       (case row_number() over (partition by rep_id order by client_registration_date ) >= 3
             when 1 then '1'
             when 2 then '2'
             else '3'
        end) as status
from reps r left join
     clients c
     on c.rep_id = r.id;

甚至：

select rep_signup_date, rep_id,  
       greatest(row_number() over (partition by rep_id order by client_registration_date ), 3) as status
from reps r left join
     clients c
     on c.rep_id = r.id;

---

谢谢，但是我添加了date\u trunc（'month'，client\u registration\u date）：：date和输出似乎每天都有：排序为rep id=1，month=Jan，status=1，然后rep id=1，month=Jan，status=2；但是我只需要status=2记录（如果是按月记录的话）。非常感谢。样本数据和期望的结果将传达您想要的信息。@GordonLinoff两者都提供了，谢谢