可以通过将ID表示为顶点并查找连接的组件来解决此问题。关于这个想法的更多信息，请参见第3.5节。让 init_table 这是你的table。首先，建立一个链接表

create table links as
select distinct id1 as v1, id2 as v2
  from init_table
 where id1 is not null and id2 is not null
union all 
select distinct id1 as v1, id3 as v2
  from init_table
 where id1 is not null and id3 is not null
union all 
select distinct id2 as v1, id3 as v2
  from init_table
 where id2 is not null and id3 is not null
;

接下来，为每个链接生成一些编号，例如行编号并执行传播：

create table links1 as
with temp_table as (
  select v1, v2, row_number() over () as score
    from links
)
, tbl1 as (
  select v1, v2, score
       , max(score) over (partition by v1) as max_1
       , max(score) over (partition by v2) as max_2
    from temp_table
)
select v1, v2, greatest(max_1, max_2) as unique_id
  from tbl1
;

然后将您的ID与匹配的表连接起来：

create table matching_table as
with temp_table as (
select v1 as id, unique_id
  from link1
union all
select v2 as id, unique_id
  from link1
)
select distinct id, unique_id
  from temp_table

如果有些ID没有耦合，那么就不难找出是哪个ID。希望这有帮助。