sql – 使用单个查询更新多个表列值

MySQL示例

MySQL中的等效代码：

UPDATE party p
LEFT JOIN party_name n ON p.party_id = n.party_id
LEFT JOIN party_details d ON p.party_id = d.party_id
LEFT JOIN incident_participant ip ON ip.party_id = p.party_id
LEFT JOIN incident i ON ip.incident_id = i.incident_id
SET
  p.employee_id = NULL,c.em_address = 'x@x.org',c.ad_postal = 'x',n.first_name = 'x',n.last_name = 'x'
WHERE
  i.confidential_dt IS NOT NULL

使用Oracle 11g会有什么相同的声明？

谢谢！

RTFM

使用Oracle时似乎单个查询不足：

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

解决方法

/** XXX CODING HORROR... */

根据您的需要,您可以使用可更新的视图.您可以创建基表的视图,并为此视图添加“代替”触发器,并直接更新视图.

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

（编辑：李大同）

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