Question

我想找到第二次（重复）的SQL记录，其值介于（15和19）之间，相隔90天（日期不需要连续）

ChildID    Value     ValuetieredDate
1           15       2015-01-01
1           16       2015-04-01
1           2        2015-05-01
2           16       2015-01-01
2           17       2015-02-01
3           15.5     2017-08-01
3           17       2017-09-07
3           18       2018-01-03
4           16       2018-01-03
4           3        2005-01-03 
5           20       2012-02-01
5           16       2012-06-02
5           19       2015-09-08

我想选择CHILDID 1,3,5 我试过了

Select distinct ChildID,b.identifier,valueTieredDate,Value
,DateDiff(Day,lead(b.valueTieredDate) OVER (Partition by b.Childid order by b.valueTieredDate desc),b.valueTieredDate) as Datedif --Find consecutive dates
,DateDiff(day,First_Value(b.valueTieredDate) Over (Partition by ChildID order by b.valueTieredDate asc),valueTieredDate ) as Datedif1 --Not consecutive  
where value between 15 and 19.9
group by ChildID,b.identifier,valueTieredDate,Value
having count(distinct value)>1
From TestTable

Answer 1

如果我弄错了，这应该有用

with cte (ChildID, Value, ValuetieredDate) as (
    select
        a, b, cast(c as datetime)
    from
        (values (1,15,'20150101')
        ,(1,16,'20150401')
        ,(1,2,' 20150501')
        ,(2,16,'20150101')
        ,(2,17,'20150201')
        ,(3,15.5,'20170801')
        ,(3,17,'20170907')
        ,(3,18,'20180103')
        ,(4,16,'20180103')
        ,(4,3,' 20050103 ')
        ,(5,20,'20120201')
        ,(5,16,'20120602')
        ,(5,19,'20150908')) t(a, b, c)
)

select 
    a.ChildID
from 
    cte a
    join cte b on a.ChildID = b.ChildID and a.ValuetieredDate < b.ValuetieredDate
where
    a.value between 15 and 19.9
    and b.value between 15 and 19.9
group by a.ChildID
having max(datediff(dd, a.ValuetieredDate, b.ValuetieredDate)) >= 90

编辑：只需从查询中删除GROUP BY

select 
    a.ChildID, a.Value, b.Value, a.ValuetieredDate, b.ValuetieredDate
from 
    cte a
    join cte b on a.ChildID = b.ChildID and a.ValuetieredDate < b.ValuetieredDate
where
    a.value between 15 and 19.9
    and b.value between 15 and 19.9
    and datediff(dd, a.ValuetieredDate, b.ValuetieredDate) >= 90

Answer 2

with cte (ChildID, Value, ValuetieredDate) as 
(   select a, b, cast(c as datetime)
    from (values (1,15,'20150101')
                ,(1,16,'20150401')
                ,(1,2,' 20150501')
                ,(2,16,'20150101')
                ,(2,17,'20150201')
                ,(3,15.5,'20170801')
                ,(3,17,'20170907')
                ,(3,18,'20180103')
                ,(4,16,'20180103')
                ,(4,3,' 20050103 ')
                ,(5,20,'20120201')
                ,(5,16,'20120602')
                ,(5,19,'20150908')
         ) t(a, b, c)
)

select distinct cte1.ChildID 
from cte as cte1 
join cte as cte2 
  on cte1.ChildID = cte2.ChildID 
 and cte1.Value between 15 and 19
 and cte2.Value between 15 and 19  
 and DATEDIFF(dd, cte1.ValuetieredDate, cte2.ValuetieredDate) >= 90

SQL用于查找值介于15和19之间的重复记录

2 个答案: