很难理解 SQL 查询

Question

我运行遍历了这个查询，但很难理解它的作用

SELECT DISTINCT
    EMPLOYEE.EMPLOYEE_ID,
    EMPLOYEE.LAST_NAME,
    EMPLOYEE.FIRST_NAME,
    COUNT(*)
FROM EMPLOYEE 
JOIN ENTRY ON EMPLOYEE.EMPLOYEE_ID = ENTRY.EMPLOYEE_ID
JOIN TICKET ON ENTRY.TICKET_ID = TICKET.TICKET_ID 
WHERE ENTRY.ACTIVITY_ID = 'ADVTS' AND EMPLOYEE.DEPARTMENT_ID ='SLS'
GROUP BY EMPLOYEE.EMPLOYEE_ID,  EMPLOYEE.LAST_NAME,
    EMPLOYEE.FIRST_NAME,ENTRY.ENTRY_ID
HAVING COUNT(ENTRY.ENTRY_ID) >= 
(SELECT CAST(1.25 * COUNT(ENTRY.ACTIVITY_ID)/COUNT(DISTINCT EMPLOYEE.EMPLOYEE_ID) AS float) 
FROM 
  EMPLOYEE 
JOIN 
  ENTRY ON EMPLOYEE.EMPLOYEE_ID = ENTRY.EMPLOYEE_ID
WHERE 
  ENTRY.ACTIVITY_ID = 'ADVTS' AND EMPLOYEE.DEPARTMENT_ID = 'SLS')

据我所知，它给出了 EMPLOYEE 的列表，他们完成了 ADVTS ACTIVITY 和来自 DEPARTMENT SLS 的 [=19] =]s 至少与 DEPARTMENT SLS 中出于 ADVTS 目的

的平均条目一样多

感谢所有花时间提供帮助的人

评论后编辑成功结果：

SELECT 
    EMPLOYEE.EMPLOYEE_ID,
    EMPLOYEE.LAST_NAME,
    EMPLOYEE.FIRST_NAME
    FROM EMPLOYEE
JOIN 
  ENTRY ON ENTRY.EMPLOYEE_ID = EMPLOYEE.EMPLOYEE_ID
GROUP 
   BY EMPLOYEE.EMPLOYEE_ID, EMPLOYEE.LAST_NAME,
    EMPLOYEE.FIRST_NAME
HAVING 
  COUNT(ENTRY.ENTRY_ID) >= 
(SELECT 
  CAST(1.25 * 
  COUNT(ENTRY.ACTIVITY_ID)/COUNT(DISTINCT EMPLOYEE.EMPLOYEE_ID)AS float) 
FROM 
  EMPLOYEE JOIN ENTRY ON EMPLOYEE.EMPLOYEE_ID = ENTRY.EMPLOYEE_ID
WHERE 
  ENTRY.ACTIVITY_ID = 'ADVTS' AND EMPLOYEE.DEPARTMENT_ID = 'SLS')

输出：

EMPLOYEE_ID| LAST_NAME| FIRST_NAME
7          | Salesman | Efficient

Answer 1

假设 TICKET_ID 在 Ticket 中是唯一的并且在 ENTRY 中永远不会是 NULL，那么你可以去掉那个 JOIN。

然后我假设查询的目的是 return 员工人数超过总体平均值的 1.25。这需要更多（合理的）假设，但这更简单地写为：

SELECT e.*
FROM (SELECT EM.EMPLOYEE_ID, EM.LAST_NAME, EM.FIRST_NAME, COUNT(*) AS CNT,   
             SUM(COUNT(*)) OVER () * 1.0 / COUNT(*) OVER () as AVG_CNT    
      FROM EMPLOYEE EM JOIN
           ENTRY EM
           ON EM.EMPLOYEE_ID = EN.EMPLOYEE_ID 
      WHERE EN.ACTIVITY_ID = 'ADVTS' AND EM.DEPARTMENT_ID = 'SLS'
      GROUP BY EM.EMPLOYEE_ID, EM.LAST_NAME, EM.FIRST_NAME
     ) e
WHERE cnt >= 1.25 * avg_cnt

很难理解 SQL 查询

Having hard time understanding SQL Query

sql

sql-server

sql-server-2019