Question

我希望显示department_id以及计数，并且计数应该超过5，我希望员工不会在1月份被录用。

我尝试了以下查询

SELECT * FROM EMPLOYEES 
WHERE DEPARTMENT_ID IN
 (
   SELECT DEPARTMENT_ID
    FROM EMPLOYEES
   GROUP BY DEPARTMENT_ID
   HAVING COUNT(*)>5 
 )
AND HIRE_DATE NOT LIKE '%JAN%';

但在这里我没有得到数数。我也想要数数。

Answer 1

如果您想要部门ID和员工数量（员工雇用日期不在1月份），那么以下内容应该有效。我说＆＃34;类似于以下内容＆＃34;因为我怀疑WHERE hire_date不喜欢'％JAN％＆＃39;可以改进，但它只取决于该列的格式。

   SELECT
      DEPARTMENT_ID, 
      COUNT(*) 
    FROM EMPLOYEES 
    WHERE HIRE_DATE NOT LIKE '%JAN%' 
    GROUP BY DEPARTMENT_ID 
    HAVING COUNT(*)>5;

如果您还希望列出各个员工以及这些部门，那么这样的事情可能有效：

SELECT a.*, b.count(*) 
FROM EMPLOYEES AS a 
INNER JOIN (   
SELECT
      DEPARTMENT_ID, 
      COUNT(*) 
    FROM EMPLOYEES 
    WHERE HIRE_DATE NOT LIKE '%JAN%' 
    GROUP BY DEPARTMENT_ID 
    HAVING COUNT(*)>5) AS b
ON a.department_id = b.department_id
WHERE a.HIRE_DATE NOT LIKE '%JAN%';

但是，我认为您可以利用您的架构来改进HIRE_DATE的where子句。一个喜欢/不喜欢的条款通常会很慢。

Answer 2

SELECT department_ID, count(employee_id) as '# of Employees' FROM EMPLOYEES 
WHERE DEPARTMENT_ID IN
 (
   SELECT DEPARTMENT_ID
    FROM EMPLOYEES
   GROUP BY DEPARTMENT_ID
   HAVING COUNT(*)>5 
 )
AND HIRE_DATE NOT LIKE '%JAN%'
group by department_ID;

此查询返回department_id，因为我按department_id分组，将返回属于每个部门的员工数

输出看起来像这样

  Department_Id | # of Employees
        1             7
        2             6
        4             9

Answer 3

从内部查询中选择计数，然后加入：

SELECT E.*, DEPT_COUNT
FROM EMPLOYEES E
JOIN (
   SELECT DEPARTMENT_ID, COUNT(*) DEPT_COUNT
   FROM EMPLOYEES
   GROUP BY DEPARTMENT_ID
   HAVING COUNT(*) > 5 
) DC ON E.DEPARTMENT_ID = DC.DEPARTMENT_ID
AND HIRE_DATE NOT LIKE '%JAN%'

员工部门明智，员工人数超过5人

3 个答案: