题解 | #给出employees表中排名为奇数行的first_name#

我的方法：两个子查询

select first_name from employees where first_name in
(select first_name from 
(select first_name,dense_rank()over(order by first_name) r from employees)t1
where r%2=1)

法二：join

SELECT
e.first_name
FROM employees e JOIN
(
SELECT 
    first_name
    , ROW_NUMBER() OVER(ORDER BY first_name ASC) AS  r_num
FROM employees
) AS t 
ON e.first_name = t.first_name
WHERE t.r_num % 2 = 1;

法三：count+自连接得出排名

select e1.first_name from employees e1
where
(
select count(*) from employees e2 where
e1.first_name>=e2.first_name
)%2 = 1