备注:测试数据库版本为MySQL 8.0
如需要scott用户下建表及录入数据语句，可参考: scott建表及录入数据sql脚本
测试数据准备:
mysql> create table new_sal(deptno int,sal int) character set utf8mb4; Query OK, 0 rows affected (0.02 sec) mysql> insert into new_sal values (10,4000); Query OK, 1 row affected (0.01 sec) mysql> select * from new_sal; +--------+------+ | deptno | sal | +--------+------+ | 10 | 4000 | +--------+------+ 1 row in set (0.00 sec)问题 用用表new_sal中的值更新表emp中相应员工的工资，条件是emp.deptno与new_sal.deptno相等，将匹配记录的emp.sal更新为new_sla.sal,将emp.comm更新为new_sal.sal的50%。
解决方案: 在表new_sal和emp之间做连接，用来查找comm新值并带给update语句。像这样通过关联子查询进行更新的情况十分常见；另一种方法是创建视图(传统视图或内联视图，视数据库支持而定)，然后更新这个视图。
update emp e inner join new_sal ns on e.deptno = ns.deptno set e.sal = ns.sal, e.comm = ns.sal/2;测试记录:
mysql> select * from emp; +-------+--------+-----------+------+------------+---------+---------+--------+ | empno | ename | job | mgr | hiredate | sal | comm | deptno | +-------+--------+-----------+------+------------+---------+---------+--------+ | 7369 | SMITH | CLERK | 7902 | 1980-12-17 | 800.00 | NULL | 20 | | 7499 | ALLEN | SALESMAN | 7698 | 1981-02-20 | 1600.00 | 300.00 | 30 | | 7521 | WARD | SALESMAN | 7698 | 1981-02-22 | 1250.00 | 500.00 | 30 | | 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975.00 | NULL | 20 | | 7654 | MARTIN | SALESMAN | 7698 | 1981-09-28 | 1250.00 | 1400.00 | 30 | | 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850.00 | NULL | 30 | | 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450.00 | NULL | 10 | | 7788 | SCOTT | ANALYST | 7566 | 1987-06-13 | 3000.00 | NULL | 20 | | 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000.00 | NULL | 10 | | 7844 | TURNER | SALESMAN | 7698 | 1981-09-08 | 1500.00 | 0.00 | 30 | | 7876 | ADAMS | CLERK | 7788 | 1987-06-13 | 1100.00 | NULL | 20 | | 7900 | JAMES | CLERK | 7698 | 1981-12-03 | 950.00 | NULL | 30 | | 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 | | 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300.00 | NULL | 10 | +-------+--------+-----------+------+------------+---------+---------+--------+ 14 rows in set (0.00 sec) mysql> update emp e -> inner join new_sal ns -> on e.deptno = ns.deptno -> set e.sal = ns.sal, -> e.comm = ns.sal/2; Query OK, 3 rows affected (0.00 sec) Rows matched: 3 Changed: 3 Warnings: 0 mysql> show warnings; Empty set (0.00 sec) mysql> mysql> select * from emp where deptno = 10; +-------+--------+-----------+------+------------+---------+---------+--------+ | empno | ename | job | mgr | hiredate | sal | comm | deptno | +-------+--------+-----------+------+------------+---------+---------+--------+ | 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 4000.00 | 2000.00 | 10 | | 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 4000.00 | 2000.00 | 10 | | 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 4000.00 | 2000.00 | 10 | +-------+--------+-----------+------+------------+---------+---------+--------+ 3 rows in set (0.00 sec)