在不使用分析功能的情况下实现排名

一尘不染

在不使用分析功能的情况下实现排名

sql

我想知道是否有一种方法可以在不使用内置函数的情况下实现SQL分析功能。

SELECT *,
    ROW_NUMBER() OVER (PARTITION BY dept_id ORDER BY salary DESC) AS rownum,
    DENSE_RANK() OVER (PARTITION BY dept_id ORDER BY salary DESC) AS denserank,
    RANK() OVER (PARTITION BY dept_id ORDER BY salary DESC) AS rnk
FROM emp;

阅读 111

2021-05-05

共1个答案

一尘不染

这是三个等效的表达式：

select emp.*,
       (select count(*)
        from emp emp2
        where emp2.dept_id = emp.dept_id and
              (emp2.salary > emp.salary or
               emp2.salary = emp.salary and emp2.emp_id <= emp.emp_id
              )
       ) as "row_number",
       (select 1 + count(*)
        from emp emp2
        where emp2.dept_id = emp.dept_id and
              emp2.salary > emp.salary 
              )
       ) as "rank",
       (select count(distinct salary)
        from emp emp2
        where emp2.dept_id = emp.dept_id and
              emp2.salary >= emp.salary
       ) as "dense_rank",
from emp;

假设存在，emp_id以使行对于“ row_number”而言是唯一的。

2021-05-05