azhou 2019-10-29
通过mysql新建teacher表如下:
mysql> create table teacher(name char(10)not null default 'xxx',age int not null default 0,salary decimal(6,1))charset=utf8; mysql> insert into teacher(name,age,salary)values('tank',18,32001.5); mysql> insert into teacher(name,age,salary)values('json',18,32001.5); mysql> insert into teacher(name,age,salary)values('nick',25,35000.7); mysql> insert into teacher(name,age,salary)values('echo',26,38000.7);
mysql> select*from teacher; +------+-----+---------+ | name | age | salary | +------+-----+---------+ | tank | 18 | 32001.5 | | json | 18 | 32001.5 | | nick | 25 | 35000.7 | | echo | 26 | 38000.7 | +------+-----+---------+ 4 rows in set (0.00 sec)
1. 查看岗位是teacher的员工姓名、年龄 2. 查看岗位是teacher且年龄大于30岁的员工姓名、年龄 3. 查看岗位是teacher且薪资在9000-1000范围内的员工姓名、年龄、薪资 4. 查看岗位描述不为NULL的员工信息 5. 查看岗位是teacher且薪资是10000或9000或30000的员工姓名、年龄、薪资 6. 查看岗位是teacher且薪资不是10000或9000或30000的员工姓名、年龄、薪资 7. 查看岗位是teacher且名字是jin开头的员工姓名、年薪
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mysql> select name,age from teacher; +------+-----+ | name | age | +------+-----+ | tank | 18 | | json | 18 | | nick | 25 | | echo | 26 | +------+-----+ 4 rows in set (0.00 sec)
mysql> select name,age from teacher where age>30; Empty set (0.00 sec)
mysql> select*from teacher where salary>1000 and salary<9000; Empty set (0.00 sec)
mysql> select*from teacher where name!=null; Empty set (0.00 sec)
mysql> select name ,age,salary from teacher where salary in(10000,9000,30000); Empty set (0.00 sec)
mysql> select name,age,salary from teacher where salary not in(10000,9000,30000); +------+-----+---------+ | name | age | salary | +------+-----+---------+ | tank | 18 | 32001.5 | | json | 18 | 32001.5 | | nick | 25 | 35000.7 | | echo | 26 | 38000.7 | +------+-----+---------+ 4 rows in set (0.00 sec)
mysql> select name,salary from teacher where name like 'jin%'; Empty set (0.00 sec)