CREATE TABLE salaries
(
salary
int(11) NOT NULL,
from_date
date NOT NULL,
to_date
date NOT NULL,
PRIMARY KEY (emp_no
,from_date
));
答案:
select s1.emp_no , s1.salary,count(distinct s2.salary) as rank
from salaries as s1,salaries as s2
where s1.to_date = '9999-01-01'
and s2.to_date = '9999-01-01'
and s1.salary <= s2.salary
group by s1.emp_no
order by s1.salary desc,s1.emp_no asc
思路:
1、count的是distinct
2、重点是 s1.salary <= s2.salary group by s1.emp_no
这样count就是rank排名,是在输出s1.salary的情况下,有多少个s2.salary大于等于s1.salary,