SQL7 查找薪水记录超过15次的员工号emp_no以及其对应的记录次数t

问题

思路

SELECT emp_no, COUNT(emp_no) as t
FROM salaries
GROUP BY emp_no
HAVING t > 15;

/*进阶思路*/

分组+聚合函数
select emp_no, count(emp_no) t
from salaries
group by emp_no
having t > 15;

补充：如果考虑第二条记录算起才算一次涨幅，应该写：

/*distinct来返回不重复字段的条数*/
select amp_no, count(distinct from_date) - 1 as t
from salaries
group by emp_no
having t > 15

补充：如果考虑严格意义上的涨幅，应该写：
select a.emp_no, count(emp_no) t
from salaries a inner join salaries b
on a.emp_no=b.emp_no and a.to_date=b.from_date
where a.salary < b.salary
group by a.emp_no
having t > 15