mysql练习1

1

查询年龄为20,21,22,23,24的女性人员信息

select * from tb_user where age in (20,21,22,23,24) and gender="女";

2

查询性别为男，年龄在20-40岁，而且姓名为三个字的员工

select * from tb_user where gender="男" and age between 20 and 40 and name like "___";

3

统计表中年龄小于60的男性和女性的数量

select gender as "性别",count(*) as "人数" from tb_user where age<60 group by gender;

4

查询年龄小于35岁的人员的姓名和年龄，并对查询结果按升序排序，如果年龄相同，则按id 降序排序

select name,age from tb_user where age<35 order by age asc,id desc;

5

查询性别为男，且年龄在20-40岁前五个人员的信息，并按年龄升序排序，年龄相同按照id升序排序

select * from tb_user where gender = "男" and age between 20 and 40 order by age asc,id asc limit 0,5;