Question

RN          LSD         ED          Aging
------------------------------------------
11111111    09-Feb-2017 06-Feb-2017  3
22222222    09-Feb-2017 02-Feb-2017  7
33333333    20-Jan-2017 29-Nov-2016 44
33333333    20-Jan-2017 07-Dec-2016 44
33333333    20-Jan-2017 29-Nov-2016 44
33333333    20-Jan-2017 30-Nov-2016 44
33333333    20-Jan-2017 29-Nov-2016 44
44444444    12-Dec-2016 19-Nov-2016 17
44444444    12-Dec-2016 19-Nov-2016 17
44444444    12-Dec-2016 25-Nov-2016 17
55555555    07-Feb-2017 04-Dec-2016 56
55555555    07-Feb-2017 04-Dec-2016 56
55555555    07-Feb-2017 13-Dec-2016 56

如何通过按aging列分组，根据最后（最多）ED Date列更新RN列？

Answer 1

我建议将日期格式更改为yyyy-mm-dd，因为它更容易计算和排序。然后你应该能够解决SQL查询的差异。

语法

DATEDIFF(datepart, startdate, enddate)

实施例

SELECT DATEDIFF(day, '2014-08-05', '2014-06-05') AS DiffDate

Answer 2

我猜你可以试试这个。但我对你的要求并不清楚。您在下面找到的代码不是解决方案中最短的代码，但可以完成这项工作。

SELECT X.RN , X.LSD, X.ED , B.Aging
FROM X 
LEFT JOIN
(
    SELECT RN , DATEDIFF(day,LSD,ED) Aging
    FROM
    (
        SELECT RN , MIN(LSD) LSD, MAX(ED) ED
        FROM X
        GROUP BY RN
    )A
)B
ON X.RN = B.RN

Answer 3

您的更新将如下所示：

const { expand, expanded } = require('expand-swagger-refs');
const schema = require('./api/swagger.json');

// Create a copy of the schema, with $ref values expanded: 
const expandedSchema = expanded(schema);

// Or expand the schema object in-place (mutates the object):
expand(schema)

但您可以使用

在视图中动态执行此操作

update tab as t1
set aging = 
  (select (DATEDIFF(day, MIN(startdate), MAX(enddate))
   from tab as t2
   where t1.rn = t2.rn
  )

Answer 4

你可以使用max和partition with CTE

;with cte as (
select *, DateDiff(day, ED, LSD) as Ag, Max(Ed) over (partition by rn) MaxEd from #yourAging
) Select a.RN, a.LSD, a.ED, Aging =  sum(tmpSm) over(partition by a.rn) 
from  ( select *,case when max(ed) over (partition by rn)=ED then Ag else 0 end as TmpSm from cte ) a

您的Code demo with input tables

Answer 5

我认为实现目标的最佳方法是像贝洛一样

DECLARE @MyTable TABLE 
(
RN int,
LSD date,
ED date,
Aging int
)

INSERT INTO @MyTable VALUES
('11111111' ,'09-Feb-2017' ,'06-Feb-2017' ,'3'),
 ('22222222' ,'09-Feb-2017' ,'02-Feb-2017' ,'7' ),
 ('33333333' ,'20-Jan-2017' ,'29-Nov-2016' ,'44'),
 ('33333333' ,'20-Jan-2017' ,'07-Dec-2016' ,'44'), 
 ('33333333' ,'20-Jan-2017' ,'29-Nov-2016' ,'44'),
 ('33333333' ,'20-Jan-2017' ,'30-Nov-2016' ,'44') ,
 ('33333333' ,'20-Jan-2017' ,'29-Nov-2016' ,'44'),
 ('44444444' ,'12-Dec-2016' ,'19-Nov-2016' ,'17') ,
 ('44444444' ,'12-Dec-2016' ,'19-Nov-2016' ,'17'),
 ('44444444' ,'12-Dec-2016' ,'25-Nov-2016' ,'17'),
 ('55555555' ,'07-Feb-2017' ,'04-Dec-2016' ,'56') ,
 ('55555555' ,'07-Feb-2017' ,'04-Dec-2016' ,'56'),
 ('55555555' ,'07-Feb-2017' ,'13-Dec-2016' ,'56')


 ;with cteToUpdate AS
 (
 SELECT RN,DATEDIFF(D,MAX(ED),MIN(LSD)) as Aging FROM @MyTable
 GROUP BY RN
 )

UPDATE source
SET source.Aging = cteToUpdate.Aging
FROM @MyTable as source
JOIN cteToUpdate ON cteToUpdate.RN =source.RN

SQL Server：基于上次日期的老化

5 个答案: