您希望所有内容都在第二个表中，然后匹配group0第一个表中的行或新行。

我认为这是join逻辑：

select coalesce(t1.group0, t2.group0) as group0, 
       coalesce(t1.group1, t2.group1) as group1,
       t1.sum_a, t2.sum_b
from table1 t1 full outer join
     table2 t2
     on t1.group0 = t2.group0 
where (t2.group0 is not null and (t1.group1 = t2.group1 or t1.group0 is null)) or
      t2.group0 is null;

使用union all以下命令更容易实现此逻辑：

select t2.group0, t2.group1, t1.sum_a, t2.sum_b
from table2 t2 left join
     table1 t1
     on t2.group0 = t1.group0 and t2.group1 = t1.group1
union all
select t1.group1, t1.group1, t1.suma, 0
from table1
where not exists (select 1 from table2 t2 where t2.group0 = t1.group0);

编辑：

修改后的问题与原始问题完全不同。那很简单full outer join：

select coalesce(t1.group0, t2.group0) as group0, 
       coalesce(t1.group1, t2.group1) as group1,
       coalesce(t1.sum_a, 0) as sum_a, coalesce(t2.sum_b, 0) as sum_b
from table1 t1 full outer join
     table2 t2
     on t1.group0 = t2.group0  and t1.group1 = t2.group1;