[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)
我有一個包含大量住宿數據的數據集(在酒店)。 每行包含開始日期和結束日期,但沒有持續時間字段。 我需要得到持續時間的總和。
樣本數據:
| 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 |
還有一些復雜的問題:
我實際上並不擅長SQL,所以我只是猜測。
Crystal Reports SQL Expression的正確語法如下所示:
(
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)
)
如果我想這樣做,這給了我一行的正確價值:
| 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 |
但我希望得到每個客戶端的持續時間的SUM,所以我想要這個:
| 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 |
我試圖在我的CASE周圍包裝一個SUM(),但這不起作用:
(
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)
)
它給出了一個錯誤,即StayDateEnd在選擇列表中無效,因為它不包含在聚合函數或GROUP BY子句中。 但我甚至不知道這意味着什么,所以我不確定如何排除故障,或者從哪里開始。 然后下一步是通過客戶端ID獲取SUM。
任何幫助將不勝感激!
您需要一個基於group by client_id的sum的子查詢以及表示子查詢之間的連接,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
雖然解釋和數據集幾乎不可能匹配,但我認為這是你想要的近似值。
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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