SQL INSERT 语句示例：员工数据插入

以下 SQL INSERT 语句示例展示了如何将员工信息插入数据库。示例数据包括员工编号、姓名、职位、经理、入职日期、薪资、佣金和部门编号。

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7369', 'SMITH', 'CLERK', '7902', '1980-12-17', '800', null, '20');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7499', 'ALLEN', 'SALESMAN', '7698', '1981-02-20', '1600', '300', '30');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7521', 'WARD', 'SALESMAN', '7698', '1981-02-22', '1250', '500', '30');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7566', 'JONES', 'MANAGER', '7839', '1981-04-02', '2975', null, '20');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7654', 'MARTIN', 'SALESMAN', '7698', '1981-09-28', '1250', '1400', '30');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7698', 'BLAKE', 'MANAGER', '7839', '1981-05-01', '2850', null, '30');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7782', 'CLARK', 'MANAGER', '7839', '1981-06-09', '2450', null, '10');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7788', 'SCOTT', 'ANALYST', '7566', '1987-06-13', '3000', null, '20');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7839', 'KING', 'PRESIDENT', null, '1981-11-17', '5000', null, '10');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7844', 'TURNER', 'SALESMAN', '7698', '1981-09-08', '1500', '0', '30');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7876', 'ADAMS', 'CLERK', '7788', '1987-06-13', '1100', null, '20');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7900', 'JAMES', 'CLERK', '7698', '1981-12-03', '950', null, '30');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7902', 'FORD', 'ANALYST', '7566', '1981-12-03', '3000', null, '20');

insert into emp (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values ('7934', 'MILLER', 'CLERK', '7782', '1982-01-23', '1300', null, '10');

错误解决：

literal does not match format string 错误通常是因为日期格式不正确导致的。请检查你的日期格式是否与数据库设置的日期格式匹配。

• 在示例中，你使用的日期格式为 'YYYY-MM-DD'。
• 如果数据库设置的日期格式为 'DD-MON-YYYY'，则你需要将日期格式更改为 'DD-MON-YYYY'。

例如，'1980-12-17' 应该更改为 '17-DEC-1980'。

注意： 数据库的日期格式设置可能因数据库类型和版本而异。请参考相关文档或咨询数据库管理员以确认正确的日期格式。