使用单个查询更新多个表列值

小编典典

使用单个查询更新多个表列值

sql

您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
左加入事件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FM

使用Oracle时，似乎只有一个查询是不够的：

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

阅读 159

2021-03-17

共1个答案

小编典典

/** XXX CODING HORROR... */

根据需要，您可以使用可 更新的视图 。您创建基本表的视图，并向该视图添加“代替”触发器，然后直接更新该视图。

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

2021-03-17