Oracle 触发器中的 RTTI
RTTI in Oracle Triggers
我有这个虚拟类型:
create or replace type Service_TY as object(
code INTEGER,
visit_analysis char(1)
)FINAL;
/
create or replace type Employee_TY as object(
dummy varchar(30)
)NOT FINAL;
/
create or replace type Doctor_TY UNDER Employee_TY(
ID INTEGER
)FINAL;
/
create or replace type Assistant_TY UNDER Employee_TY(
ID INTEGER
)FINAL;
/
create or replace type Habilitation_TY as object(
employee ref Employee_TY,
service ref Service_TY
)FINAL;
/
还有这些虚拟表:
CREATE TABLE Service of Service_TY(
code primary key,
visit_analysis not null check (visit_analysis in ('v', 'a'))
);
/
CREATE TABLE Doctor of Doctor_TY(
ID primary key
);
/
CREATE TABLE Assistant of Assistant_TY(
ID primary key
);
/
CREATE TABLE Habilitation of Habilitation_TY;
/
我想创建一个触发器,当在 Habilitation 中插入一个新的元组时,应该检查如果员工是助理(而不是医生),visit_analysis 属性等于 'a' 知道它是否是一个合法的元组。
不知道如何查看Employee的类型(是医生还是助理)
我会这样做:
create or replace
TRIGGER CHECK_HABILITATION
BEFORE INSERT ON HABILITATION
FOR EACH ROW
DECLARE
BEGIN
IF (:NEW.EMPLOYEE is of ASSISTANT_TY)
THEN
IF :NEW.SERVICE.visit_analysis = 'v'
THEN
raise_application_error(-10000, 'invalid tuple');
END IF;
END;
但它不起作用。
我应该如何检查该类型?
我得到的错误是:
Error(14,4): PLS-00103: Encountered the symbol ";" when expecting one of the following: if
根据 IS OF
条件的文档,您需要将类型括在括号中,例如:
IF (:NEW.EMPLOYEE is of (ASSISTANT_TY) )
每 https://docs.oracle.com/cd/B28359_01/server.111/b28286/conditions014.htm#SQLRF52157.
我不太熟悉对象类型的使用,因此可能还有其他一些我没有发现的问题。
尝试放到一个变量中,下面这个应该可以。
create or replace
TRIGGER CHECK_HABILITATION
BEFORE INSERT ON HABILITATION
FOR EACH ROW
DECLARE
emp employee_TY;
ser service_TY;
BEGIN
select deref(:new.employee) into emp from dual;
if (emp is of (assistant_ty)) then
select deref(:new.service) into ser from dual;
if ser.visit_analysis = 'v' then
raise_application_error('-20001', 'invalid tuple');
end if;
end if;
END;
/
我有这个虚拟类型:
create or replace type Service_TY as object(
code INTEGER,
visit_analysis char(1)
)FINAL;
/
create or replace type Employee_TY as object(
dummy varchar(30)
)NOT FINAL;
/
create or replace type Doctor_TY UNDER Employee_TY(
ID INTEGER
)FINAL;
/
create or replace type Assistant_TY UNDER Employee_TY(
ID INTEGER
)FINAL;
/
create or replace type Habilitation_TY as object(
employee ref Employee_TY,
service ref Service_TY
)FINAL;
/
还有这些虚拟表:
CREATE TABLE Service of Service_TY(
code primary key,
visit_analysis not null check (visit_analysis in ('v', 'a'))
);
/
CREATE TABLE Doctor of Doctor_TY(
ID primary key
);
/
CREATE TABLE Assistant of Assistant_TY(
ID primary key
);
/
CREATE TABLE Habilitation of Habilitation_TY;
/
我想创建一个触发器,当在 Habilitation 中插入一个新的元组时,应该检查如果员工是助理(而不是医生),visit_analysis 属性等于 'a' 知道它是否是一个合法的元组。
不知道如何查看Employee的类型(是医生还是助理)
我会这样做:
create or replace
TRIGGER CHECK_HABILITATION
BEFORE INSERT ON HABILITATION
FOR EACH ROW
DECLARE
BEGIN
IF (:NEW.EMPLOYEE is of ASSISTANT_TY)
THEN
IF :NEW.SERVICE.visit_analysis = 'v'
THEN
raise_application_error(-10000, 'invalid tuple');
END IF;
END;
但它不起作用。
我应该如何检查该类型?
我得到的错误是:
Error(14,4): PLS-00103: Encountered the symbol ";" when expecting one of the following: if
根据 IS OF
条件的文档,您需要将类型括在括号中,例如:
IF (:NEW.EMPLOYEE is of (ASSISTANT_TY) )
每 https://docs.oracle.com/cd/B28359_01/server.111/b28286/conditions014.htm#SQLRF52157.
我不太熟悉对象类型的使用,因此可能还有其他一些我没有发现的问题。
尝试放到一个变量中,下面这个应该可以。
create or replace
TRIGGER CHECK_HABILITATION
BEFORE INSERT ON HABILITATION
FOR EACH ROW
DECLARE
emp employee_TY;
ser service_TY;
BEGIN
select deref(:new.employee) into emp from dual;
if (emp is of (assistant_ty)) then
select deref(:new.service) into ser from dual;
if ser.visit_analysis = 'v' then
raise_application_error('-20001', 'invalid tuple');
end if;
end if;
END;
/