根据您的示例，假设id2始终存在，您可以聚合行，按id2分组：

select max(id1) id1,  id2, max(id3) id3 from
( --your dataset as in example
 select 'A'  as id1, 1 as id2,  null   as id3 union all
 select null as id1, 1 as id2, 'Alpha' as id3 union all
 select null as id1, 2 as id2, 'Beta'  as id3 union all
 select 'A'  as id1, 2 as id2,  null   as id3
 )s
 group by id2;

OK
A       1       Alpha
A       2       Beta
Time taken: 58.739 seconds, Fetched: 2 row(s)

现在我试着实现你描述的逻辑：

select --pass2
 id1, id2, id3,
 case when lag(id2) over (order by id2, GroupId) = id2 then lag(GroupId) over (order by id2, GroupId) else GroupId end GroupId2
 from
 (
 select        --pass1
 id1, id2, id3,
 case when 
 lag(id1) over(order by id1, NVL(ID1,ID3)) =id1 then lag(NVL(ID1,ID3))  over(order by id1, NVL(ID1,ID3)) else NVL(ID1,ID3) end GroupId
 from
( --your dataset as in example
 select 'A'  as id1, 1 as id2,  null   as id3 union all
 select null as id1, 1 as id2, 'Alpha' as id3 union all
 select null as id1, 2 as id2, 'Beta'  as id3 union all
 select 'A'  as id1, 2 as id2,  null   as id3
 )s
 )s --pass1
;

OK
id1     id2     id3     groupid2
A       1       NULL    A
NULL    1       Alpha   A
A       2       NULL    A
NULL    2       Beta    A
Time taken: 106.944 seconds, Fetched: 4 row(s)