题解 | 给出employees表中排名为奇数行的first_name

with t1 as (
    select first_name
        , rank() over (order by first_name) as sort_rank
    from employees
    order by first_name
), t2 as (
    select first_name
    from t1
    where t1.sort_rank %2!=0
)
select e.first_name
from employees e
join t2 on t2.first_name = e.first_name 
;
#第一步：使用窗口函数计算每个用户的排名
#第二步：筛选出排名为奇数的所有用户
#第三步：观察答案，他是需要按照原表的顺序显示用户的first_name，联接表即可