我正在尝试计算代表有多少时间让x客户申请服务：这意味着我需要date_created之间的时间 - 即。约会被加入的日期，以及当代表达到某个“状态”时。当rep的客户端（=用户）的x具有非空date_applied时，即达到状态 - 即。约会用户注册。
x是达到每个“状态”的最低标准，并且与之前的问题相关联：Aggregate case when inside non aggregate query我正在计算“状态”，如下所示：
case when count(date_applied) over (partition by rep_id) >=10 then 'status1'
when count(date_applied) over (partition by rep_id) >=5 then 'status2'
when count(date_applied) over (partition by rep_id) >=1 then 'status3'
else 'no_status' end status
因此，需要10个客户才能达到status1，5个达到status2，1达到status3。这些是每个“状态”的标准，因此，如果您有7个用户，则仍然会根据第5个用户应用的日期计算status2。
我认为计算time_to_status1/2/3（我想要得到的）应该是这样的：
case when count(date_applied) over (partition by rep_id) >=10 then
datediff(day, date_created, date_applied for the 10th user that applied with that rep) end as time_to_status1,
case when count(date_applied) over (partition by rep_id) >=5 then
datediff(day, date_created, date_applied for the 5th user that applied with that rep) end as time_to_status2,
case when count(date_applied) over (partition by rep_id) >=1 then
datediff(day, date_created, date_applied for the 1st user that applied with that rep) end as time_to_status3
非常感谢任何帮助！
- 编辑 -
示例当前数据：
rep_id user_id date_created date_applied status
1 1 1/1/2018 6:43:22 AM 1/5/2018 2:45:15 PM status2
1 2 1/1/2018 6:43:22 AM 1/5/2018 3:35:15 PM status2
1 3 1/1/2018 6:43:22 AM 1/6/2018 4:25:15 PM status2
1 4 1/1/2018 6:43:22 AM 1/7/2018 5:05:15 PM status2
1 5 1/1/2018 6:43:22 AM 1/10/2018 3:35:15 PM status2
1 6 1/1/2018 6:43:22 AM 1/15/2018 12:55:23 PM status2
2 7 1/12/2018 1:13:42 PM 1/15/2018 4:25:15 PM status3
2 8 1/12/2018 1:13:42 PM 1/16/2018 1:05:15 PM status3
2 9 1/12/2018 1:13:42 PM 1/16/2018 3:35:15 PM status3
3 10 1/20/2018 10:13:15 AM 1/26/2018 7:25:15 PM status3
4 11 1/21/2018 3:33:23 PM (null) no_status
期望的输出：
rep_id user_id date_created date_applied status time_to_status1 time_to_status2 time_to_status3
1 1 1/1/2018 6:43:22 AM 1/5/2018 2:45:15 PM status2 (null) 9 (null)
1 2 1/1/2018 6:43:22 AM 1/5/2018 3:35:15 PM status2 (null) 9 (null)
1 3 1/1/2018 6:43:22 AM 1/6/2018 4:25:15 PM status2 (null) 9 (null)
1 4 1/1/2018 6:43:22 AM 1/7/2018 5:05:15 PM status2 (null) 9 (null)
1 5 1/1/2018 6:43:22 AM 1/10/2018 3:35:15 PM status2 (null) 9 (null)
1 6 1/1/2018 6:43:22 AM 1/15/2018 12:55:23 PM status2 (null) 9 (null)
2 7 1/12/2018 1:13:42 PM 1/15/2018 4:25:15 PM status3 (null) (null) 3
2 8 1/12/2018 1:13:42 PM 1/16/2018 1:05:15 PM status3 (null) (null) 3
2 9 1/12/2018 1:13:42 PM 1/16/2018 3:35:15 PM status3 (null) (null) 3
3 10 1/20/2018 10:13:15 AM 1/26/2018 7:25:15 PM status3 (null) (null) 6
4 11 1/21/2018 3:33:23 PM (null) no_status (null) (null) (null)
rep_id=1有status2，因为他有6个用户使用非空date_applied，因此time_to_status2在他的情况下基于第5个客户的date_applied代表注册：datediff(day, '1/1/2018 6:43:22 AM', '1/10/2018 3:35:15 PM') = 9 天
rep_id=2有status3，因为他有3个非空date_applied的用户，因此time_to_status3在他的情况下基于第一个客户代表的date_applied已注册：datediff(day, '1/12/2018 1:13:42 PM', '1/15/2018 4:25:15 PM') = 3 天
rep_id=3有status3，因为他有1个（＆gt; = 1）非空date_applied的用户，因此time_to_status3就是datediff(day, '1/20/2018 10:13:15 AM', '1/26/2018 7:25:15 PM') = 6 天
答案 0 :(得分：0)
基于@ Parfait删除的提示，以及@ Gordon对另一个问题的回答，我能够得出答案：
with cte as
(
initial query with:
case when count(client_signup_date) over (partition by rep_id) >=10 then 'status1'
when count(client_signup_date) over (partition by rep_id) >=5 then 'status2'
when count(client_signup_date) over (partition by rep_id) >=1 then 'status3'
else 'none' end status,
row_number() over(partition by rep_id order by client_signup_date) as rank
)
select *,
max(case when status = 'status1' and rank = 10
then datediff(day, advisor_onboard_date, client_signup_date)
end) over (partition by rep_id) as time_to_status1,
max(case when status = 'status2' and rank = 5
then datediff(day, advisor_onboard_date, client_signup_date)
end) over (partition by rep_id) as time_to_status2,
max(case when status = 'status3' and rank = 1
then datediff(day, advisor_onboard_date, client_signup_date)
end) over (partition by rep_id) as time_to_status3
into #t
from cte