SQL Oracle - 最佳查询答案

【问题标题】：SQL Oracle - optimum querySQL Oracle - 最佳查询
【发布时间】：2015-03-23 10:14:16
【问题描述】：

我有三个具有以下架构的表：

Persons(Person_id, Department_id, Sure_name, Name, Birthyear, Height, Manager_id)
Departments (Department_id, Department_Name, Code)
Salaries (Salary_id, Person_id, Salary)

我需要运行一个查询，该查询将显示部门名称，其中工人的最小和最大高度之间的差异最大。

我是通过以下方式完成的：

select Department_Name
from Departments
where Department_id = (select Department_id
                      from Departments 
                      join Persons 
                      using (Department_id)
                      group by Department_id
                      having max(height) - min(height) = (select max(max(height) - min(height)) 
                                                          from Departments 
                                                          join Persons 
                                                          using (Department_id) 
                                                          group by Department_id));

而且效果很好，只是我不确定解决方案是否最佳，这里有两个嵌套查询，我想知道我是否可以用更简单的方式实现。

【问题讨论】：

标签： sql oracle performance join

【解决方案1】：

试试这个。

SELECT Department_Name
FROM   Departments D
       INNER JOIN (SELECT Department_id,
                          Max(height) - Min(height) AS diff
                   FROM   Departments
                          JOIN Persons
                            ON using (Department_id)
                   WHERE ROWNUM = 1
                   GROUP  BY Department_id
                   ORDER  BY diff DESC) B
               ON d.Department_id = b.Department_id

或使用Window Function

SELECT Department_id,
       Department_Name
FROM   (SELECT Row_number()OVER(ORDER BY Max(height)- Min(height) desc ) rn,
               Department_id,
               Department_Name
        FROM   Departments
               JOIN Persons
                 ON using(Department_id)
        GROUP  BY Department_id,
                  Department_Name)a
WHERE  rn = 1

【讨论】：

您甚至可以另外按部门名称分组以省略外部连接
内部联接是否等同于联接？另外，我认为“限制”不适用于 Oracle。
@user4359659 - Join 等于 Inner Join。添加了另一个答案
我不太确定可以将聚合放在窗口函数的OVER( ) 子句中。你不会想PARTITION BY department_id。
@DavidFaber - 是的 partition by 这里不需要。据我所知，您可以在 order by 中使用聚合

【解决方案2】：

这可能对你有用，有点复杂，但使用分析函数 RANK() 可能比只使用聚合更简单：

SELECT d.department_id, d.department_name, d.code, p1.height_diff
  FROM departments d INNER JOIN (
    SELECT department_id, height_diff, RANK() OVER ( ORDER BY height_diff DESC ) AS rn
      FROM (
        SELECT department_id, MAX(height) - MIN(height) AS height_diff
          FROM Persons
         GROUP BY department_id
      )
  ) p1
    ON d.department_id = p1.department_id
 WHERE p1.rn = 1;

Please see SQL Fiddle demo here. 注意我之前点错了，现在更正为DESC。

【讨论】：

正在编辑...缺少右括号。