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假设我有以下表结构:

a (a_id number primary key, a_code varchar unique not null)
b (b_id number primary key, a_id number foreign key references a(a_id), b_value varchar

如果它们尚不存在,我需要在表上插入b几百对,并且我拥有的信息是. 我来自 Postgres 背景,这样做是微不足道的,但我很难找到一种在 Oracle 中实现它的方法,而无需求助于创建并随后删除物化视图,或者每行插入一个。(a_id, b_value)(a_code, b_value)

这是我能想出的最不难看的解决方案:

insert /*+ IGNORE_ROW_ON_DUPKEY_INDEX(b_value, iun_b_value) */
into b (b_id, a_id, b_value)
select b_id.nextval, a_id, b_value
from a
inner join

(select 'code1' a_code, 'value1' b_value from dual union all
select 'code2' a_code, 'value2' b_value from dual union all
select 'code3' a_code, 'value3' b_value from dual union all
select 'code4' a_code, 'value4' b_value from dual) new

on a.a_code = new.b_value

但是我被告知我不能使用该语义提示来处理生产中的重复行,所以我尝试了这个:

merge into b using
  (select a_id, b_value
  from a inner join 

  (select 'code1' a_code, 'value1' b_value from dual union all
  select 'code2' a_code, 'value2' b_value from dual union all
  select 'code3' a_code, 'value3' b_value from dual union all
  select 'code4' a_code, 'value4' b_value from dual) new

  new on a.a_code = new.a_code) insert_codes

on (insert_codes.a_code = b.a_code and insert_codes.b_value = b.b_value)
when not matched then
  insert (b.b_id, b.a_id, b.b_value)
  values (b_id.nextval, insert_codes.a_id, 1, insert_codes.b_value)

但是当我尝试使用子句"ORA-00980: synonym translation is no longer valid"解决此问题时,我在子选择中引用别名时遇到了同样的错误。where not exists

有没有办法在没有物化视图或多个插入的情况下做到这一点?

4

1 回答 1

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查看我的简化示例让我意识到出了什么问题。该on条款是

on (insert_codes.a_code = b.a_code and insert_codes.b_value = b.b_value)

但应该是

on (insert_codes.a_id = b.a_id and insert_codes.b_value = b.b_value)

所以这有效:

merge into b using
  (select a_id, b_value
  from a inner join 

  (select 'code1' a_code, 'value1' b_value from dual union all
  select 'code2' a_code, 'value2' b_value from dual union all
  select 'code3' a_code, 'value3' b_value from dual union all
  select 'code4' a_code, 'value4' b_value from dual) new

  new on a.a_code = new.a_code) insert_codes

on (insert_codes.a_id = b.a_id and insert_codes.b_value = b.b_value)
when not matched then
  insert (b.b_id, b.a_id, b.b_value)
  values (b_id.nextval, insert_codes.a_id, 1, insert_codes.b_value)

在这里发帖,以防其他人将来无法完成相同类型的插入。

于 2017-03-14T12:53:28.697 回答