Question

我有两张桌子：

表A：

ID    NAME
1     First
2     Second
3     Second
4     Third
5     First

表B（引用表A记录）：

A_ID    REC
1       10
2       10
3       20
4       20
5       20
3       30

表A有2个重复记录，我可以轻松删除，但我需要用较少的重复记录ID更新表B.我想最终得到：

A_ID    REC
1       10
2       10
2       20
4       20
1       20
2       30

我已经为MySQL解决了这个问题，但由于Oracle不支持UPDATE语句中的连接，因此卡在了Oracle上。

Answer 1

您可以将此作为查询执行。首先，获得新的id：

select a1.id, min(a2.id) as min_id
from a a1 left join
     a a2
     on a1.name = a2.name
group by a1.id;

然后，作为查询：

select b.*, a.min_id
from b join
     (select a1.id, min(a2.id) as min_id
      from a a1 left join
           a a2
           on a1.name = a2.name
      group by a1.id
     ) a
     on b.a_id = a.id;

如果你真的需要更新，那有点棘手：

update b
    set b.a_id = (select min(a2.id) as min_id
                  from a a1 left join
                       a a2
                       on a1.name = a2.name
                  where a1.id = b.a_id
                 )
    where b.id <> (select min(a2.id) as min_id
                   from a a1 left join
                        a a2
                        on a1.name = a2.name
                   where a1.id = b.a_id
                  )

更新Oracle中引用表的重复记录

1 个答案: