MySQL - 错误的值

Question

我尝试计算一些数字，我一直发誓，因为它失败了，然后我尝试了这个:

SELECT SUM( 10 *1 ) 
FROM user_achievements
INNER JOIN achievements
WHERE user_achievements.user_id =8

上面写着:420 ?!?

我尝试让它工作:

SELECT SUM((SELECT score_base FROM achievements WHERE id = user_achievements.achievement_id)*((SELECT pixels_multiplier FROM achievements WHERE id = user_achievements.achievement_id)) * achievement_level) * achievement_level FROM user_achievements INNER JOIN achievements WHERE user_achievements.user_id=2

成就: ID， 水平， 像素基数， 分数基础， 像素乘数

用户成就: 用户身份， 成就_id， 成就级别

Answer 1

如果您的查询返回420，则意味着聚合前结果集中返回了42行。

这是您的查询:

SELECT sum(10*1)
FROM user_achievements ua cross join
     achievements a
WHERE ua.user_id = 8;

如果我不得不猜测，你错过了一个连接:

SELECT sum(10*1)
FROM user_achievements ua join
     achievements a
     on ua.achievement_id = a.id
WHERE ua.user_id = 8;

编辑:

这是评论中的查询:

SELECT sum((SELECT score_base
            FROM achievements
            WHERE id = ua.achievement_id
            ) *
            (SELECT pixels_multiplier
             FROM achievements
             WHERE id = ua.achievement_id
            )
           )
FROM user_achievements ua join
     achievements a
     on ua.achievement_id = a.id
WHERE ua.user_id = 2;

它甚至不应该解析。聚合中不允许使用子查询。试试这个:

SELECT sum(a.score_base * a.pixels_multiplier)
FROM user_achievements ua join
     achievements a
     on ua.achievement_id = a.id
WHERE ua.user_id = 2;