Question

很难创建触发器。这是我的代码。我创建了2个表：

CREATE TABLE client_table
  (
  clientid NUMBER(6) PRIMARY KEY,
  lastname varchar2(50) NOT NULL,
  firstname varchar2(50) NOT NULL,
  password varchar2(20) NOT NULL,
  email varchar2(100) NOT NULL,
  phoneno NUMBER(12) NOT NULL,
  zipcode NUMBER(6) NOT NULL ,
  street varchar2(60) NOT NULL,
  CHECK (email LIKE '%@%.%')
  );

CREATE TABLE location_table
(
  zipcode NUMBER(6) PRIMARY KEY,
  country varchar2(100) NOT NULL,
  region varchar2(60) NOT NULL,
  town varchar2(60) NOT NULL
  );

我添加了一个约束，以便client_table中的zipcode是外键。

ALTER TABLE client_table
ADD CONSTRAINT clirefzip
  FOREIGN KEY(zipcode) REFERENCES location_table(zipcode)
  DEFERRABLE INITIALLY DEFERRED;

现在我想添加一个触发器，这样当我从location_table中删除一个位置时，它将从client_table中删除所有与删除位置相同的邮政编码的客户端。所以我尝试这样的事情：

CREATE OR REPLACE TRIGGER location_delete
AFTER DELETE ON location_table
REFERENCING OLD AS old_buffer
FOR EACH ROW
DELETE FROM client_table WHERE zipcode = old_buffer.zipcode;

不幸的是它不起作用。

Answer 1

Hi there has some syntactical errors in your code. I hope this code works . Thanks    

    CREATE OR REPLACE TRIGGER location_delete
    AFTER DELETE ON location_table
    --REFERENCING OLD AS old_buffer
    FOR EACH ROW
    begin
    DELETE FROM client_table WHERE zipcode = zipcode; 
    end location_trigger;

在Oracle中创建触发器时出错

1 个答案: