结果重复

Question

下面的代码是重复我的部门名称结果，我不知道为什么。我需要显示部门的名称，在那里工作的员工的工作ID和工资。另外，我需要弄清楚是否在CUBE表达式中使用了department_id或job_id。

SELECT d.department_name AS "Name", 
       e.job_id AS "Job ID", 
       SUM(e.salary) AS "salary", 
       DECODE(GROUPING(d.department_id),1, 'Yes', 'No') AS "Department Used", 
       DECODE(GROUPING(e.job_id), 1, 'Yes', 'No') AS "Job used"
FROM employees e 
    full OUTER JOIN departments d 
    ON (e.department_id = d.department_id)
GROUP BY CUBE(d.department_name, e.job_id, d.department_id)
ORDER BY d.department_name;

Answer 1

因为cube会为所有组合生成小计。在您的示例中，组合的工作方式如下：

1）department_name - job_id - department_id

2）department_name - job_id

3）department_name - department_id

4）department_name

5）job_id - department_id

6）job_id

7）department_id

8）（总计）

这就是你获得重复值的原因。

我不知道你到底需要什么，但我认为你需要这样的东西：

  select d.department_name as "Name",
         e.job_id as "Job ID",
         sum(e.salary) as "Salary",
         decode(grouping(d.department_id), 1, 'Yes', 'No') as "Department Used",
         decode(grouping(e.job_id), 1, 'Yes', 'No') as "Job Used"
    from hr.employees e
    full outer join hr.departments d
      on (e.department_id = d.department_id)
   group by cube(e.job_id), d.department_name, d.department_id
   order by d.department_name;

Answer 2

问题在于您的CUBE()子句：您在DEPARTMENT中包含两个属性，因此对于DEPARTMENT_ID和DEPARTMENT_NAME的所有组合，结果都是立方体。

然而，这两列是相关的：ID和NAME之间存在一对一的关系（假设您正在使用HR模式或其他一些不错的数据模型）。​​

因此解决方案非常简单：将CUBE()子句更改为使用一个。由于DEPARTMENT_NAME正在进行预测，因此需要选择。您需要更改DECODE()以匹配。

SELECT d.department_name AS "Name", 
       e.job_id AS "Job ID", 
       SUM(e.salary) AS "salary", 
       DECODE(GROUPING(d.department_name),1, 'Yes', 'No') AS "Department Used", 
       DECODE(GROUPING(e.job_id), 1, 'Yes', 'No') AS "Job used"
FROM employees e 
    full OUTER JOIN departments d 
    ON (e.department_id = d.department_id)
GROUP BY CUBE(d.department_name, e.job_id)
ORDER BY d.department_name;