I'm struggling to find a solution to update (not just select) SQL table records based on previous values. I can get the script to populate the immediate subsequent record with the LAG() function, but when I have the same values in a row (like 'CC' below) I can't get it to populate with the next value that's not the same as the current value.
我很难找到一个解决方案,根据以前的值更新(而不仅仅是选择)SQL表记录。我可以使用LAG()函数来获取脚本以填充紧接的后续记录,但是当我在一行中具有相同的值(如下面的“CC”)时,我无法使用下一个值来填充它不是与当前值相同。
It would also be helpful to be able to add CASE/WHEN condition so that only values with the same BaseID are evaluated. Any help would be greatly appreciated.
能够添加CASE / WHEN条件以便仅评估具有相同BaseID的值也是有帮助的。任何帮助将不胜感激。
Here is my desired result:
这是我想要的结果:
BaseID Value Date NextValue
1 AA 2017-10-01 BB
1 BB 2017-10-02 CC
1 CC 2017-10-03 DD
1 CC 2017-10-03 DD
1 CC 2017-10-03 DD
1 DD 2017-10-04 NULL
2 EE 2017-10-01 FF
2 FF 2017-10-02 GG
2 GG 2017-10-03 NULL
2 个解决方案
#1
0
Get the distinct baseid,value,date combinations and use lead
to get the next value in a cte and use itto update
.
获取不同的基数,值,日期组合并使用线索获取cte中的下一个值并使用它进行更新。
with cte as (select t1.baseid,t1.value,t1.nextvalue,t2.nxt_value
from tbl t1
left join (select t.*,lead(value) over(partition by baseid order by datecol) as nxt_value
from (select distinct baseid,datecol,value from tbl) t
) t2
on t1.baseid=t2.baseid and t1.datecol=t2.datecol and t1.value=t2.value
)
update cte set nextvalue=nxt_value
This assumes there can't be multiple values for a given baseid,date combination.
这假设给定的baseid,date组合不能有多个值。
#2
0
Here is a working example using DENSE_RANK as another option.
这是一个使用DENSE_RANK作为另一个选项的工作示例。
declare @Something table
(
BaseID int
, MyValue char(2)
, MyDate date
, NextValue char(2)
)
insert @Something
(
BaseID
, MyValue
, MyDate
) VALUES
(1, 'AA', '2017-10-01')
, (1, 'BB', '2017-10-02')
, (1, 'CC', '2017-10-03')
, (1, 'CC', '2017-10-03')
, (1, 'CC', '2017-10-03')
, (1, 'DD', '2017-10-04')
, (2, 'EE', '2017-10-01')
, (2, 'FF', '2017-10-02')
, (2, 'GG', '2017-10-03')
;
with SortedResults as
(
select *
, DENSE_RANK() over(partition by BaseID order by BaseID, MyDate ) as MyRank
from @Something
)
update sr set NextValue = sr2.MyValue
from SortedResults sr
join SortedResults sr2 on sr2.MyRank - 1 = sr.MyRank and sr.BaseID = sr2.BaseID
select *
from @Something
#1
0
Get the distinct baseid,value,date combinations and use lead
to get the next value in a cte and use itto update
.
获取不同的基数,值,日期组合并使用线索获取cte中的下一个值并使用它进行更新。
with cte as (select t1.baseid,t1.value,t1.nextvalue,t2.nxt_value
from tbl t1
left join (select t.*,lead(value) over(partition by baseid order by datecol) as nxt_value
from (select distinct baseid,datecol,value from tbl) t
) t2
on t1.baseid=t2.baseid and t1.datecol=t2.datecol and t1.value=t2.value
)
update cte set nextvalue=nxt_value
This assumes there can't be multiple values for a given baseid,date combination.
这假设给定的baseid,date组合不能有多个值。
#2
0
Here is a working example using DENSE_RANK as another option.
这是一个使用DENSE_RANK作为另一个选项的工作示例。
declare @Something table
(
BaseID int
, MyValue char(2)
, MyDate date
, NextValue char(2)
)
insert @Something
(
BaseID
, MyValue
, MyDate
) VALUES
(1, 'AA', '2017-10-01')
, (1, 'BB', '2017-10-02')
, (1, 'CC', '2017-10-03')
, (1, 'CC', '2017-10-03')
, (1, 'CC', '2017-10-03')
, (1, 'DD', '2017-10-04')
, (2, 'EE', '2017-10-01')
, (2, 'FF', '2017-10-02')
, (2, 'GG', '2017-10-03')
;
with SortedResults as
(
select *
, DENSE_RANK() over(partition by BaseID order by BaseID, MyDate ) as MyRank
from @Something
)
update sr set NextValue = sr2.MyValue
from SortedResults sr
join SortedResults sr2 on sr2.MyRank - 1 = sr.MyRank and sr.BaseID = sr2.BaseID
select *
from @Something