我想合并一组记录
(id) / (referencedid)
1 10
1 11
2 11
2 10
3 10
3 11
3 12
查询结果应为
1 10
1 11
3 10
3 11
3 12
因此,由于id = 1且id = 2具有相同的一组相应的referenceids {10,11},因此它们将被合并。但是id = 3 s对应的referenceid是不一样的,因此不会被合并。
完成这项工作的好方法是什么?
答案 0 :(得分:1)
Select id, referenceid
From MyTable
Where Id In (
Select Min( Z.Id ) As Id
From (
Select Z1.id, Group_Concat( Z1.referenceid ) As signature
From (
Select id, referenceid
From MyTable
Order By id, referenceid
) As Z1
Group By Z1.id
) As Z
Group By Z.Signature
)
答案 1 :(得分:0)
-- generate count of elements for each distinct id
with Counts as (
select
id,
count(1) as ReferenceCount
from
tblReferences R
group by
R.id
)
-- generate every pairing of two different id's, along with
-- their counts, and how many are equivalent between the two
,Pairings as (
select
R1.id as id1
,R2.id as id2
,C1.ReferenceCount as count1
,C2.ReferenceCount as count2
,sum(case when R1.referenceid = R2.referenceid then 1 else 0 end) as samecount
from
tblReferences R1 join Counts C1 on R1.id = C1.id
cross join
tblReferences R2 join Counts C2 on R2.id = C2.id
where
R1.id < R2.id
group by
R1.id, C1.ReferenceCount, R2.id, C2.ReferenceCount
)
-- generate the list of ids that are safe to remove by picking
-- out any id's that have the same number of matches, and same
-- size of list, which means their reference lists are identical.
-- since id2 > id, we can safely remove id2 as a copy of id, and
-- the smallest id of which all id2 > id are copies will be left
,RemovableIds as (
select
distinct id2 as id
from
Pairings P
where
P.count1 = P.count2 and P.count1 = P.samecount
)
-- validate the results by just selecting to see which id's
-- will be removed. can also include id in the query above
-- to see which id was identified as the copy
select id from RemovableIds R
-- comment out `select` above and uncomment `delete` below to
-- remove the records after verifying they are correct!
--delete from tblReferences where id in (select id from RemovableIds) R