select post,group_concat(name),count(id) from employee group by post having count(id)<2; select post, avg(salary) from employee group up post having avg(salary)>10000 and avg(salary)<20000;
order by
select * from employee order by age asc; #默认升序 select * from employee order by age desc; select * from employee order by age asc,id desc; select post, count(id) as emp_count from employee where salary>500 group by post having count(id)>1 order by count(id) desc;
原文:https://www.cnblogs.com/zhujing666/p/12313947.html