MySQL-错误的值
我试着计算一些数字,我一直发誓,因为它失败了,然后我试着这样做:MySQL-错误的值,mysql,sql,Mysql,Sql,我试着计算一些数字,我一直发誓,因为它失败了,然后我试着这样做: SELECT SUM( 10 *1 ) FROM user_achievements INNER JOIN achievements WHERE user_achievements.user_id =8 上面写着:420 我试着让它起作用: SELECT SUM((SELECT score_base FROM achievements WHERE id = user_achievements.achievement_id)*(
SELECT SUM( 10 *1 )
FROM user_achievements
INNER JOIN achievements
WHERE user_achievements.user_id =8
上面写着:420
我试着让它起作用:
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:
身份证件
水平,
像素作为基础,
得分基础,
像素/倍增器
用户成就:
用户id,
成就(id),
如果查询为return 420,则表示在聚合之前的结果集中返回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;
如果让我猜,您缺少一个连接:
SELECT sum(10*1)
FROM user_achievements ua join
achievements a
on ua.achievement_id = a.id
WHERE ua.user_id = 8;
编辑:
这是注释中的查询:
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;
1054-条款“中的未知列a.a.U id”只需将a更改为ua,将ua更改为a即可comparison@user3249998 . . . 我不知道您表中字段的名称。我正在展示一个查询的示例。将其用作给定数据库的指南。我尝试过此操作,但未起作用:选择sumSELECT score\u base FROM magnities,其中id=user\u sagnities.magnities\u id*选择pixels\u multiplier FROM magnities,其中id=user\u sagnities.magnities\u id FROM user\u sagnities用户\u sagnities加入成就a上的成就user_acgregations.acgregation_id=a.id,其中user_acgregations.user_id=2;两者都给出了这个错误:1054-未知列“user\u acgregations.user\u id”在“where子句”中可以显示这两个表的create语句。您可以省略任何未使用的字段。已添加!希望有帮助: