Question

我想选择工作场所中年龄最大和最年轻的员工。年龄最大和年龄最小的是他们的工作经验。我提到了SQL Server的先前问题，并如下修改了我的代码，但出现错误：

无效的标识符。

我想知道我对更改代码有什么建议吗？

 SELECT d.Dtable_name,
        RANK() OVER(ORDER BY SUM(ROUND((SYSDATE - e.hire_date) / 365.24,0)) DESC) AS yrsexperience
   FROM Dtable_department d
   LEFT JOIN Etable_employee e
     ON d.department_id=e.department_id
WHERE yrsexperience =(SELECT MAX(d.department_name) KEEP(DENSE_RANK FIRST ORDER BY SUM(ROUND((SYSDATE - e.hire_date) / 365.24,0)) DESC) AS yrsexperience) AS Oldest)FROM Etable_employee e)

OR yrsexperience =(SELECT
       MIN(d.department_name) KEEP(DENSE_RANK LAST  ORDER BY SUM(ROUND((SYSDATE - e.hire_date) / 365.24,0)) DESC) AS yrsexperience) AS Youngest
  FROM Etable_employee e)

更新

我参考下面的答案后编辑了代码，但它抛出了错误：

找不到FROM。

Answer 1

您可以使用FIRST和LAST聚合函数。

SELECT MAX(name) KEEP(DENSE_RANK FIRST ORDER BY yrs_experience ) AS Youngest,
       MAX(name) KEEP(DENSE_RANK LAST  ORDER BY yrs_experience ) AS Oldest
  FROM etable_employee;

Answer 2

看来YRSEXPERIENCE是ETABLE_EMPLOYEE上的一个字段。鉴于此，我希望如此

SELECT *
  FROM ETABLE_EMPLOYEE
  WHERE YRS_EXPERIENCE IN (SELECT MIN(YRS_EXPERIENCE) FROM ETABLE_EMPLOYEE
                           UNION ALL
                           SELECT MAX(YRS_EXPERIENCE) FROM ETABLE_EMPLOYEE)

将为您提供所需的内容。

编辑

根据更多信息：

SELECT SYSDATE - e.HIRE_DATE AS YRS_EXPERIENCE
  FROM ETABLE_EMPLOYEE e
  WHERE SYSDATE - e.HIRE_DATE IN (SELECT MIN(SYSDATE - HIRE_DATE) FROM ETABLE_EMPLOYEE
                                  UNION ALL
                                  SELECT MAX(SYSDATE - HIRE_DATE) FROM ETABLE_EMPLOYEE)

dbfiddle here

选择Oracle中最老和最年轻的员工

更新

2 个答案:

编辑