3

您将如何使用单个查询更新多个表中的数据?

MySQL 示例

MySQL中的等效代码:

更新方 p
左加入party_name n ON p.party_id = n.party_id
左加入party_details d ON p.party_id = d.party_id
左加入 event_participant ip ON ip.party_id = p.party_id
LEFT JOIN 事件 i ON ip.incident_id = i.incident_id
放
  p.employee_id = NULL,
  c.em_address = 'x@x.org',
  c.ad_postal = 'x',
  n.first_name = 'x',
  n.last_name = 'x'
在哪里
  i.confidential_dt 不为空

使用 Oracle 11g 的相同语句是什么?

谢谢!

实时调频

使用 Oracle 时,单个查询似乎是不够的:

http://download-west.oracle.com/docs/cd/B10501_01/server.920/a96540/statements_108a.htm#2067717

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4 回答 4

5
/** XXX CODING HORROR... */

根据您的需要,您可以使用可更新的视图。您创建基表的视图并向该视图添加“替代”触发器,然后直接更新视图。

一些示例表:

create table party (
    party_id integer,
    employee_id integer
    );

create table party_name (
    party_id integer,
    first_name varchar2(120 char),
    last_name varchar2(120 char)
    );

insert into party values (1,1000);   
insert into party values (2,2000);
insert into party values (3,3000);

insert into party_name values (1,'Kipper','Family');
insert into party_name values (2,'Biff','Family');
insert into party_name values (3,'Chip','Family');

commit;

select * from party_v;

PARTY_ID    EMPLOYEE_ID    FIRST_NAME    LAST_NAME
1            1000           Kipper        Family
2            2000           Biff          Family
3            3000           Chip          Family

...然后创建一个可更新的视图

create or replace view party_v
as
select
    p.party_id,
    p.employee_id,
    n.first_name,
    n.last_name
from
    party p left join party_name n on p.party_id = n.party_id;

create or replace trigger trg_party_update
instead of update on party_v 
for each row
declare
begin
--
    update party
    set
        party_id = :new.party_id,
        employee_id = :new.employee_id
    where
        party_id = :old.party_id;
--
    update party_name
    set
        party_id = :new.party_id,
        first_name = :new.first_name,
        last_name = :new.last_name
    where
        party_id = :old.party_id;
--
end;
/

您现在可以直接更新视图...

update party_v
set
    employee_id = 42,
    last_name = 'Oxford'
where
    party_id = 1;

select * from party_v;

PARTY_ID    EMPLOYEE_ID    FIRST_NAME    LAST_NAME
1            42             Kipper        Oxford
2            2000           Biff          Family
3            3000           Chip          Family
于 2010-04-23T09:43:24.220 回答
1

我遇到了同样的问题,我在 Oracle 中找不到简单的方法来做到这一点。

看这里: Oracle 更新声明了解更多信息。

于 2010-04-23T00:21:12.033 回答
1

您可以使用 OracleMERGE语句来执行此操作。它是一种基于将目标表与内联视图连接起来的批量更新或插入语句。

MERGE INTO bonuses D
   USING (
      SELECT employee_id, salary, department_id FROM employees
      WHERE department_id = 80
   ) S ON (D.employee_id = S.employee_id)
 WHEN MATCHED THEN 
   UPDATE SET D.bonus = D.bonus + S.salary*.01
 WHEN NOT MATCHED THEN 
   INSERT (D.employee_id, D.bonus)
   VALUES (S.employee_id, S.salary*0.1);

如果您不需要插入部分,则只需省略上面的最后 3 行。

于 2010-04-23T04:20:40.340 回答
1

在某些情况下,可以使用 PL/SQL 来实现这一点。在我的例子中,我通过一些标准在两个表中搜索匹配的行,然后在循环中更新每一行。

像这样的东西:

begin
  for r in (
    select t1.id as t1_id, t2.id as t2_id
    from t1, t2
    where ...
  ) loop
    update t1
    set ...
    where t1.id = r.t1_id;

    update t2
    set ...
    where t2.id = r.t2_id;
  end loop;
end;
于 2015-11-30T11:33:14.550 回答