Question

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分析函数。有人可以帮我弄这个吗。谢谢。

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;

Answer 1

以下是三个等效表达式：

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;

这假设存在emp_id以使行对“row_number”唯一。

Answer 2

您可以使用相关的子查询执行此操作。

select dept_id,salary,
(select count(*) from emp e1 where e1.dept_id=e.dept_id and e1.salary>=e.salary) as rnum
from emp e

当没有关系时，这很有效。

Answer 3

这适用于所有情况

select DEPT_ID,SALARY,

(select count(*)+1  from emp r where r.SALARY>o.SALARY and r.dept_id=o.dept_id) **rank**,


(select count(distinct SALARY )+1  from emp r where r.SALARY>o.SALARY and r.dept_id=o.dept_id) *d_rank*,



(select count(*)+1  from (select x.*,rownum rn from ( select emp.* from emp  order by DEPT_ID asc,salary desc ) x) r where r.rn<o.rn and r.dept_id=o.dept_id) **rownumm**


from (select x.*,rownum rn from ( select emp.* from emp  order by DEPT_ID asc,salary desc ) x) o 

order by DEPT_ID,salary desc;

排名：-使用（（计数（小于当前行的值）+1）

对于密集排名：-与排名相同（计数不同值小于当前行）+1

row_number：-通过为每行生成rownum来创建嵌套查询，该行对于所有行都是不同的。现在最重要的是与等级逻辑相同（大于上一个rownum（选择子查询的行数）的值的数量）+1

不使用分析函数实现Rank

3 个答案: