使用完全联接覆盖表。 FULL JOIN 返回联接记录+未从左表联接+未从右表联接。您可以使用case语句实现如下逻辑：

insert OVERWRITE table test1

select 
      --select t1 if both or t1 only exist, t2 if only t2 exists
      case when t1.ID is null then t2.ID   else t1.ID   end as ID,
      case when t1.ID is null then t2.NAME else t1.NAME end as NAME,
      case when t1.ID is null then t2.CODE else t1.CODE end as CODE,

      --if found in t1 but not in t2 then current_date else leave as is
      case when (t1.ID is not null) and (t2.ID is null) then current_date else t1.CORRECTED_DATE end as CORRECTED_DATE 
  from test1 t1 
       FULL OUTER JOIN test2 t2 on t1.ID=t2.ID;

另请参见有关增量更新的类似问题，您的逻辑不同，但方法相同：https://stackoverflow.com/a/37744071/2700344
使用数据进行测试：

with test1 as (
select stack (2,
1, 'TEST',    3,null,    
29,'TEST2',   90 , null
             ) as (ID,NAME,CODE,CORRECTED_DATE)
),

     test2 as (
select stack (2,
              12,'TEST5',20,
              1,'TEST',3
             ) as (ID, NAME, CODE)
)

select 
      --select t1 if both or t1 only exist, t2 if only t2 exists
      case when t1.ID is null then t2.ID   else t1.ID   end as ID,
      case when t1.ID is null then t2.NAME else t1.NAME end as NAME,
      case when t1.ID is null then t2.CODE else t1.CODE end as CODE,

      --if found in test1 but not in test2 then current_date else leave as is
      case when (t1.ID is not null) and (t2.ID is null) then current_date else t1.CORRECTED_DATE end as CORRECTED_DATE 
  from test1 t1 
       FULL OUTER JOIN test2 t2 on t1.ID=t2.ID;

结果：

OK
id      name    code    corrected_date
1       TEST    3       NULL
12      TEST5   20      NULL
29      TEST2   90      2019-03-14
Time taken: 41.727 seconds, Fetched: 3 row(s)

结果如预期。