day 34 作业

作业

mysql> create table tea(
    -> id int unsigned auto_increment primary key,
    -> name varchar(50),
    -> age int,
    -> salary int,
    -> desclib varchar(50) default null
    -> )charset=utf8;
    
insert into tea(name,age,salary) values ('Jackie','19','9000');
insert into tea(name,age,salary) values ('Coach','32','30000');
insert into tea(name,age,salary) values ('Andy','26','15000');
insert into tea(name,age,salary) values ('jinx','16','3500');

# 1. 查看岗位是teacher的员工姓名、年龄
select name,age from tea;

# 2. 查看岗位是teacher且年龄大于30岁的员工姓名、年龄
select name,age from tea where age>30;

# 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
select name,age,salary from tea where salary between 1000 and 9000;

# 4. 查看岗位描述不为NULL的员工信息
select * from tea where desclib is not null;

# 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
select name,age,salary from tea where salary in (10000,9000,30000)

# 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
select name,age,salary from tea where salary not in (10000,9000,30000);

# 7. 查看岗位是teacher且名字是jin开头的员工姓名、年薪
select name,salary*12 from tea where name like 'jin%';
posted @ 2019-10-29 20:20  colacheng  阅读(...)  评论(...编辑  收藏
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