数据库查询SQL_MySQL

　现在要统计每个age中，高、中、和低级别的数目。即通过age作group by，但在每个age值内，还要按level取值统计各级别的记录数。

解决方案：

select age, (select count(*) from testtbl t2 where t2.age=t1.age and level

(select count(*) from testtbl t2 where t2.age=t1.age and level =?)as mid,

(select count(*) from testtbl t2 where t2.age=t1.age and level >?)as high

from testtbl t1 group by age;