Question

我有四张桌子用户，工资，奖金，罚款我想选择薪水，包括一些用户的所有奖金和罚款我在每个表中都有user_id（工资，奖金，罚款）我怎样才能做到这一点可以有两个或更多的奖金和罚金我试着做

BEGIN
   SELECT SUM(salary.salary) + SUM(bonus.bonus) - SUM(penalty.penalty) FROM users
   INNER JOIN salary on salary.user_id = users.id 
   INNER JOIN bonus on bonus.user_id = users.id
   INNER JOIN penalty on penalty.user_id = users.id;
   END

但是每次奖金和罚款都有多个结果我想只有一个号码谢谢你的帮助

Answer 1

假设给定用户可以在每个salary，bonus和penalty表中有多个记录，这里的一个安全方法是在单独的子查询中汇总金额，然后将结果连接在一起：

SELECT
    u.id,
    COALESCE(s.salary, 0) + COALESCE(b.bonus, 0) - COALESCE(p.penalty, 0) AS amount
FROM users u
LEFT JOIN
(
    SELECT user_id, SUM(salary) AS salary
    FROM salary
    GROUP BY user_id
) s
    ON u.id = s.user_id
LEFT JOIN
(
    SELECT user_id, SUM(bonus) AS bonus
    FROM bonus
    GROUP BY user_id
) b
    ON u.id = b.user_id
LEFT JOIN
(
    SELECT user_id, SUM(penalty) AS penalty
    FROM penalty
    GROUP BY user_id
) p
    ON u.id = p.user_id

您可以在上面添加WHERE子句以限制一个或多个用户，例如

WHERE u.id = some_value

薪水选择与奖金mysql

1 个答案: