[我正在尝试编写一个触发器,将触发器的先前版本存储在给定名为tablename的表中的audit_tablename表中。
这里是代码...
CREATE OR REPLACE FUNCTION process_ui_audit()
RETURNS TRIGGER AS
$$
DECLARE
audit_table_name text := TG_TABLE_SCHEMA || '.audit_' || TG_TABLE_NAME;
audit_table_schema text := TG_TABLE_SCHEMA;
BEGIN
IF (TG_OP = 'UPDATE')
THEN
EXECUTE FORMAT('INSERT INTO %1$I SELECT NEXTVAL(''$1.hibernate_sequence''),now(), user, ($1).*',
audit_table_name, audit_table_schema)
USING OLD;
NEW.version = OLD.version + 1;
RETURN NEW;
ELSIF (TG_OP = 'INSERT')
THEN
NEW.version = 1;
RETURN NEW;
END IF;
END;
[当我尝试更新一行时,触发器运行,并且出现类似这样的错误...。
[42P01]错误:关系“ webapp.audit_portal_user”不存在其中:PL / pgSQL函数webapp.process_ui_audit()第13行位于EXECUTE
我不知道我是不正确地格式化表名还是其他?表名webapp.audit_portal_user明确存在。
它无需指定架构名称即可工作。
这里是一个简化的示例:
create table portal_user(
uid int,
uname text
);
CREATE TABLE
create table audit_portal_user(
uid int,
uname text,
who text,
what text,
ts timestamp
);
CREATE TABLE
create or replace function process_ui_audit()
returns trigger as
$$
declare
audit_table_name text := 'audit_' || tg_table_name;
begin
if (tg_op = 'UPDATE')
then
execute format('insert into %I values($1.*, user, %L, now())',
audit_table_name, 'UPDATE') using new;
return null;
end if;
end;
$$
language plpgsql;
CREATE FUNCTION
create trigger audit
after update on portal_user
for each row
execute function process_ui_audit();
CREATE TRIGGER
insert into portal_user values(12, 'titi');
INSERT 0 1
select * from portal_user;
uid | uname
-----+-------
12 | titi
(1 row)
update portal_user set uname='toto' where uid=12;
UPDATE 1
select * from portal_user;
uid | uname
-----+-------
12 | toto
(1 row)
select * from audit_portal_user;
uid | uname | who | what | ts
-----+-------+----------+--------+----------------------------
12 | toto | postgres | UPDATE | 2020-06-01 10:20:36.549257
(1 row)