用SQL写出: 求每一年龄上人数超过2的男生的具体人数,并按年龄从小到大...

having count(*)>2

having count(major)>2

select 性别,case when 年龄 between 20 and 29 then 1 else 0 end [20-29],case when 年龄 between 30 and 39 then 1 else 0 end [30-39],case when 年龄 between 40 and 49 then 1 else 0 end [40-49]from 表名group by 性别 以上使用于大部分数据库 只是在起别名上，只适用于sq...

select COUNT(Ssex)sex,AVG(Sage)age from Student group by Ssex

select floor(年龄/10) as age from 表 )group by age 这样基本效果就出来了，达到楼主的要求就要加如函数计算了 sql语法 select convert(varchar，age*10)+'--'+convert(varchar，(age+1)*10) ,count(age) from (select floor(年龄/10) as age from 表 )group by age oracle语法 select ...

select 系别,性别,count(*) 人数 from table group by 系别,性别 order by 人数 desc 上面语句不行的话：select * from (select 系别,性别,count(*) 人数 from table group by 系别,性别) order by 人数 desc

select user_age 年龄,count(user_age) 人数,cast(cast((count(user_age)/((select count(*) from worker)*1.0)*100) as decimal(9,2)) as varchar)+'%' 所占比例 from worker group by user_age

1 select count(*) as 人数,性别,部分号 from 员工表 group by 性别,部门号 having 人数>2 and 性别=男;2 select 部门 from 部门表 where 部门号 in (select 部门号 from (select count(*) as 人数,性别,部分号 from 员工表 group by 性别,部门号 having 人数>2 and 性别=男));

'赵%' union all select avg(age) as 平均年龄 from tablename union all select min(age) as 最小年龄 from tablename union all select max(age) as 最大年龄 from tablename 执行以上SQL语句，我们可以得到一个包含张、王、李、赵姓学生人数以及平均年龄、最小年龄、最大年龄的结果集。

select '30-35岁' as 年龄段 count(*) as 人数 from tb where year(getdate())-year(birthday)>=30 and year(getdate())-year(birthday)<35 union all select '35-40岁' as 年龄段 count(*) as 人数 from tb where year(getdate())-year(birthday)>=35 and year(getdate())-year(...