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我有如下结构的表 p_it_people 并在其上创建了触发器。

CREATE TABLE  "P_IT_PEOPLE" 
   (    "PERSON_ID" NUMBER NOT NULL ENABLE, 
    "PERSON_NAME" VARCHAR2(255) NOT NULL ENABLE, 
    "PERSON_EMAIL" VARCHAR2(255) NOT NULL ENABLE, 
    "PERSON_ROLE" VARCHAR2(30) NOT NULL ENABLE, 
    "USERNAME" VARCHAR2(255) NOT NULL ENABLE, 
    "ASSIGNED_DEPT" NUMBER, 
    "CREATED_ON" DATE NOT NULL ENABLE, 
    "CREATED_BY" VARCHAR2(255) NOT NULL ENABLE, 
    "MODIFIED_ON" DATE, 
    "MODIFIED_BY" VARCHAR2(255), 
    "PERSON_PASSWORD" VARCHAR2(100), 
    "APPROVER" VARCHAR2(50), 
     CONSTRAINT "P_IT_PEOPLE_PK" PRIMARY KEY ("PERSON_ID")
  USING INDEX  ENABLE, 
     CONSTRAINT "P_IT_PEOPLE_NAME_UK" UNIQUE ("PERSON_NAME")
  USING INDEX  ENABLE, 
     CONSTRAINT "P_IT_PEOPLE_USERNAME_UK" UNIQUE ("USERNAME")

表上的现有触发器：

CREATE OR REPLACE EDITIONABLE TRIGGER  "P_IT_PEOPLE_BIU" 
before insert or update on p_it_people
for each row
begin
if inserting then
if :NEW.PERSON_ID is null then
:NEW.PERSON_ID := it_api.gen_pk;
end if;
:NEW.CREATED_ON := sysdate;
:NEW.CREATED_BY := nvl(v('APP_USER'),USER);
end if;
if updating then
:NEW.MODIFIED_ON := sysdate;
:NEW.MODIFIED_BY := nvl(v('APP_USER'),USER);
end if;
end;

除此之外，我想创建另一个触发器，它会在创建新条目时发送电子邮件。我部署了这个触发器：

CREATE OR REPLACE EDITIONABLE TRIGGER  "P_IT_ISSUES_AIU_NEW_PASSWORD" 
AFTER 
insert  on P_IT_PEOPLE
for each row 
DECLARE 
v_person_id number;
v_username varchar2(50);
v_Email varchar2(255);
Begin
select person_id,username,person_email into v_person_id,v_username,v_email from p_it_people where person_id=v_person_id;

             APEX_MAIL.SEND( 
                 p_to => v_email, 
                 p_from => v_email, 
                 p_body =>  'Your account has been created  ' ||chr(10)|| 
                 ' Username'|| v_username||chr(10)|| 
                 ' Password'||v_username  ,
                  p_subj => 'New User'); 

end;

现在，当我尝试插入行时，它会抛出错误-p_it_people 正在变异。我该如何反击？

4

1 回答 1

1

您正在从同一个表中选择，因为行级触发器正在触发 - 正是导致“表正在变异”的原因。但在这种情况下，你甚至不需要。只需像这样使用 :NEW 伪记录：

CREATE OR REPLACE EDITIONABLE TRIGGER  "P_IT_ISSUES_AIU_NEW_PASSWORD" 
AFTER 
insert  on P_IT_PEOPLE
for each row 
Begin
             APEX_MAIL.SEND( 
                 p_to => :NEW.person_email,
                 p_from => :NEW.person_email , 
                 p_body =>  'Your account has been created  ' ||chr(10)|| 
                 ' Username'|| :NEW.username||chr(10)|| 
                 ' Password'||:NEW.username  ,
                  p_subj => 'New User'); 

end;
于 2020-04-28T14:48:30.937 回答