Question

我需要找到在我的数据库中收到奖金的用户。我感兴趣的唯一用户是那些获得奖金超过一次的用户。

我应该如何使用此查询才能获得多次获得奖励的用户？

select Bonus, BonusUser, BonusType, Amount 
from Bonus 
where BonusType="1" 
order by BonusUser asc;

我需要一个提示所有“重复”行的查询，这样我就可以从中删除奖励。

我之前没有解释过，但是有些用户利用了一个bug并且可以获得免费奖金，所以我必须选择那些重复的行，分析并删除它是否是滥用案例。

Answer 1

你可以这样做

   select BonusUser,
          count(*)
   from Bonus 
  where BonusType="1" 
  group by BonusUser
   having count(*)>1
  order by BonusUser asc

你必须提供一些预期结果的虚拟数据。

Answer 2

添加GROUP BY和HAVING子句

SELECT Bonus
    , BonusUser
    , BonusType
    , Ammount 
FROM Bonus 
WHERE BonusType="1" 
GROUP BY BonusUser
HAVING Count(*) > 1
ORDER BY BonusUser asc;

根据您的评论，我认为这就是您想要的，这将为您提供所有有奖金的用户列表，但它会为您提供有多个奖金的人数：

SELECT Bonus
    , t.BonusUser
    , BonusType
    , amount 
    , t2.cntbonus
FROM Bonus t
inner join 
(
    select count(*) as CntBonus, bonususer
    from Bonus
    where BonusType='1'
    group by bonususer
) t2
    on t.bonususer = t2.bonususer
WHERE BonusType='1'
ORDER BY BonusUser asc

Answer 3

作为其他答案的替代方案，您还可以执行以下操作：

SELECT DISTINCT b1.*
FROM Bonus b1
JOIN Bonus b2
  ON b1.BonusUser = b2.BonusUser
  AND b1.Id > b2.Id
WHERE b1.BonusType = "1"
  AND b2.BonusType = "1"
ORDER BY b1.BonusUser ASC;

选择表中仅重复的行

3 个答案: