力扣1076(MySQL)-员工项目Ⅱ（简单）

题目：

编写一个SQL查询，报告所有雇员最多的项目。

查询结果格式如下所示：

 

 解题思路：

方法一：将两个表联结，以project_id进行分组，统计员工数降序排序，然后筛选出第一条数据。

select project_id
from project a
join employee b
on a.employee_id = b.employee_id
group by project_id
order by count(a.employee_id) desc
limit 1

 方法二:子查询，先查询出最大数，然后在用主查询的统计数等于子查询的最大数。

select project_id
from project_1075 a
group by project_id
having count(employee_id) = (
   select count(*) as cout
   from project_1075
   group by project_id
   order by cout desc
   limit 1
)