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[英]How to get a Datediff calculation based on two different case statements?
[英]SQL: Get an aggregate (SUM) of a calculation of two fields (DATEDIFF) that has conditional logic (CASE WHEN)
我有一个包含大量住宿数据的数据集(在酒店)。 每行包含开始日期和结束日期,但没有持续时间字段。 我需要得到持续时间的总和。
样本数据:
| Stay ID | Client ID | Start Date | End Date |
| 1 | 38 | 01/01/2018 | 01/31/2019 |
| 2 | 16 | 01/03/2019 | 01/07/2019 |
| 3 | 27 | 01/10/2019 | 01/12/2019 |
| 4 | 27 | 05/15/2019 | NULL |
| 5 | 38 | 05/17/2019 | NULL |
还有一些复杂的问题:
我实际上并不擅长SQL,所以我只是猜测。
Crystal Reports SQL Expression的正确语法如下所示:
(
SELECT (CASE
WHEN StayDateStart < DATEADD(year,-1,CURRENT_TIMESTAMP) THEN DATEDIFF(day,DATEADD(year,-1,CURRENT_TIMESTAMP),ISNULL(StayDateEnd,CURRENT_TIMESTAMP))
ELSE DATEDIFF(day,StayDateStart,ISNULL(StayDateEnd,CURRENT_TIMESTAMP))
END)
)
如果我想这样做,这给了我一行的正确价值:
| Stay ID | Client ID | Start Date | End Date | Duration |
| 1 | 38 | 01/01/2018 | 01/31/2019 | 210 | // only days since June 4 2018 are counted
| 2 | 16 | 01/03/2019 | 01/07/2019 | 4 |
| 3 | 27 | 01/10/2019 | 01/12/2019 | 2 |
| 4 | 27 | 05/15/2019 | NULL | 21 |
| 5 | 38 | 05/17/2019 | NULL | 19 |
但我希望得到每个客户端的持续时间的SUM,所以我想要这个:
| Stay ID | Client ID | Start Date | End Date | Duration |
| 1 | 38 | 01/01/2018 | 01/31/2019 | 229 | // 210+19
| 2 | 16 | 01/03/2019 | 01/07/2019 | 4 |
| 3 | 27 | 01/10/2019 | 01/12/2019 | 23 | // 2+21
| 4 | 27 | 05/15/2019 | NULL | 23 |
| 5 | 38 | 05/17/2019 | NULL | 229 |
我试图在我的CASE周围包装一个SUM(),但这不起作用:
(
SELECT SUM(CASE
WHEN StayDateStart < DATEADD(year,-1,CURRENT_TIMESTAMP) THEN DATEDIFF(day,DATEADD(year,-1,CURRENT_TIMESTAMP),ISNULL(StayDateEnd,CURRENT_TIMESTAMP))
ELSE DATEDIFF(day,StayDateStart,ISNULL(StayDateEnd,CURRENT_TIMESTAMP))
END)
)
它给出了一个错误,即StayDateEnd在选择列表中无效,因为它不包含在聚合函数或GROUP BY子句中。 但我甚至不知道这意味着什么,所以我不确定如何排除故障,或者从哪里开始。 然后下一步是通过客户端ID获取SUM。
任何帮助将不胜感激!
您需要一个基于group by client_id的sum的子查询以及表示子查询之间的连接,例如:
select Stay_id, client_id, Start_date, End_date, t.sum_duration
from your_table
inner join (
select Client_id,
SUM(CASE
WHEN StayDateStart < DATEADD(year,-1,CURRENT_TIMESTAMP) THEN DATEDIFF(day,DATEADD(year,-1,CURRENT_TIMESTAMP),ISNULL(StayDateEnd,CURRENT_TIMESTAMP))
ELSE DATEDIFF(day,StayDateStart,ISNULL(StayDateEnd,CURRENT_TIMESTAMP))
END) sum_duration
from your_table
group by Client_id
) t on t.Client_id = your_table.client_id
虽然解释和数据集几乎不可能匹配,但我认为这是你想要的近似值。
declare @your_data table (StayId int, ClientId int, StartDate date, EndDate date)
insert into @your_data values
(1,38,'2018-01-01','2019-01-31'),
(2,16,'2019-01-03','2019-01-07'),
(3,27,'2019-01-10','2019-01-12'),
(4,27,'2019-05-15',NULL),
(5,38,'2019-05-17',NULL)
;with data as (
select *,
datediff(day,
case
when datediff(day,StartDate,getdate())>365 then dateadd(year,-1,getdate())
else StartDate
end,
isnull(EndDate,getdate())
) days
from @your_data
)
select *,
sum(days) over (partition by ClientId)
from data
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