Question

我有这个查询

  SELECT salary
    FROM worker W
    JOIN single_user U ON u.users_id_user = W.single_user_users_id_user
    JOIN university_has_single_user US ON US.single_user_users_id_user = U.users_id_user
    JOIN course C ON C.id_course = US.course_id_course
    JOIN formation_area FA ON FA.id_formation_area = C.formation_area_id_formation_area
   WHERE FA.area = "Multimédia"
GROUP BY users_id_user

......给出了这个输出：

salary
--------
1400.00
800.00

如何计算此输出的平均值？如果我添加：

SELECT round(avg (salary), 0)

...输出再次是1400.00和800.00，而不是平均值（因为分组依据）。

Answer 1

使用：

 SELECT AVG(DISTINCT salary)
   FROM worker W
   JOIN single_user U ON u.users_id_user = W.single_user_users_id_user
   JOIN university_has_single_user US ON US.single_user_users_id_user = U.users_id_user
   JOIN course C ON C.id_course = US.course_id_course
   JOIN formation_area FA ON FA.id_formation_area = C.formation_area_id_formation_area
  WHERE FA.area = "Multimédia"

由于salary列未包含在聚合中，per the documentation, the values you see are arbitrary (can't be guaranteed 100% of the time)。

通常，您需要一个派生表来获取不同值的平均值，但MySQL's AVG supports using DISTINCT within it。

查询中的平均值 - mysql

1 个答案: