你可以 cross join 名单 userid s和 clothesid 生成所有组合，然后使用 not exists 在原始表上标识缺少的行：

select u.userid, c.clothesid
from (select distinct userid from mytable) u
cross join (select distinct clothesid from mytable) c
where not exists(
    select 1 from mytable t on t.userid = u.userid and t.clothesid = c.clothesid
)

我想你想要：

select (case when t1.clothesid is not null then 2 else 1 end),
       coalesce(t1.clothesid, t2.clothesid)
from (select t.*
      from t
      where t.userid = 1
     ) t1 full join
     (select t.*
      from t
      where t.userid = 2
     ) t2
     on t1.clothesid = t2.clothesid
where t1.clothesid is null or t2.clothesid is null;

实际上，我想我有一个更简单的解决办法：

select (case when min(t.userid) = 1 then 2 else 1 end), clothesid
from t
group by clothesid
having count(*) = 1;

这是一把小提琴。

左键连接所有 userid 以及 clothesid 返回表并仅返回不匹配的行：

select t1.userid, t2.clothesid
from (select distinct userid from tablename) t1
cross join (select distinct clothesid from tablename) t2
left join tablename t on t.userid = t1.userid and t.clothesid = t2.clothesid
where t.userid is null

或者和接线员一起 EXCEPT :

select t1.userid, t2.clothesid
from (select distinct userid from tablename) t1
cross join (select distinct clothesid from tablename) t2
except
select userid, clothesid
from tablename

请看演示。
结果：

> userid | clothesid
> -----: | --------:
>      1 |         4
>      1 |         5
>      2 |         3