oracle 创建表时出现“缺少右括号”错误

6yjfywim  于 2022-11-03  发布在  Oracle
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我一直收到错误“ORA-00907:缺少右括号”时运行以下create table语句:

create table employee(
  primary key(emp_id number(20)),
  emp_name varchar(30),
  birth_date date CHECK(birth_date>18),
  gender varchar(10),
  dept_no number(20)
    CONSTRAINT fk FOREIGN KEY(dept_no)
    REFERENCES department(dept_no),
  address varchar(50),
  designation varchar(20)
    CHECK(designation IN('manager', 'clerk', 'leader', 'analyst', 'designer', 'coder','tester')),
  salary number(50)
    CHECK(salary>0),
  experience number(2),
  email_id varchar(30)
    CONSTRAINT chk_email
    CHECK (REGEXP_LIKE(email_id,'^[A-Za-z0-9_.]+@[A-Za-z]+\.[A-Za-z]{2,4}$'))
);

我已经查了确切的语法,检查了很多次,一切似乎都很完美,但错误仍然存在。什么是错的?

juzqafwq

juzqafwq1#

一点点的

  • 语法无效(主键关键字的位置),
  • 多余的外键关键字(您应该 * 行外 * 使用它们,而不是 * 行内 *),
  • birth_date列的检查约束条件错误(日期怎么会大于18?),
  • Oracle建议我们使用varchar2而不是varchar
  • number(50)的精度太高(也许您宁愿跳过它)。

修复后(使用虚拟主表):

SQL> CREATE TABLE department
  2  (
  3     dept_no   NUMBER PRIMARY KEY
  4  );

Table created.

员工姓名:

SQL> CREATE TABLE employee
  2  (
  3     emp_id        NUMBER (20) PRIMARY KEY,
  4     emp_name      VARCHAR2 (30),
  5     birth_date    DATE,
  6     gender        VARCHAR2 (10),
  7     dept_no       NUMBER CONSTRAINT fk_emp_dept REFERENCES department (dept_no),
  8     address       VARCHAR2 (50),
  9     designation   VARCHAR2 (20)
 10                     CHECK
 11                        (designation IN ('manager',
 12                                         'clerk',
 13                                         'leader',
 14                                         'analyst',
 15                                         'designer',
 16                                         'coder',
 17                                         'tester')),
 18     salary        NUMBER CHECK (salary > 0),
 19     experience    NUMBER (2),
 20     email_id      VARCHAR2 (30)
 21                     CONSTRAINT chk_email CHECK
 22                        (REGEXP_LIKE (
 23                            email_id,
 24                            '^[A-Za-z0-9_.]+@[A-Za-z]+\.[A-Za-z]{2,4}$'))
 25  );

Table created.

SQL>

作为检查员工年龄的触发器,以下是方法:

SQL> CREATE OR REPLACE TRIGGER trg_bi_emp
  2     BEFORE INSERT
  3     ON employee
  4     FOR EACH ROW
  5  BEGIN
  6     IF MONTHS_BETWEEN (SYSDATE, :new.birth_date) < 18 * 12
  7     THEN
  8        raise_application_error (-20000,
  9                                 'Too young; must be at least 18 years of age');
 10     END IF;
 11  END;
 12  /

Trigger created.

SQL> INSERT INTO employee (emp_id, birth_date) VALUES (1, DATE '2020-07-25');
INSERT INTO employee (emp_id, birth_date) VALUES (1, DATE '2020-07-25')
            *
ERROR at line 1:
ORA-20000: Too young; must be at least 18 years of age
ORA-06512: at "SCOTT.TRG_BI_EMP", line 4
ORA-04088: error during execution of trigger 'SCOTT.TRG_BI_EMP'

SQL> INSERT INTO employee (emp_id, birth_date) VALUES (1, DATE '1997-07-25');

1 row created.

SQL>

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