所有员工的当前(to_date='9999-01-01')薪水按照salary进行按照1-N的排名

不使用窗口函数

-- rank排名：查询表中大于自己薪水的员工的数量（考虑并列：去重）
SELECT 
  s1.emp_no,
  s1.salary,
  (SELECT 
    COUNT(DISTINCT s2.salary) 
  FROM
    salaries s2 
  WHERE s2.to_date = '9999-01-01' 
    AND s2.salary >= s1.salary) AS `rank`  -- 去重：计算并列排名
FROM
  salaries s1 
WHERE s1.to_date = '9999-01-01' 
ORDER BY s1.salary DESC,
  s1.emp_no ;

使用窗口函数

SELECT 
  emp_no,
  salary,
  dense_rank () over (
ORDER BY salary DESC) AS `rank` 
FROM
  salaries 
WHERE to_date = '9999-01-01' ;