如何只选择那些有多个记录的用户？

我有以下查询。我想调整为仅选择具有多个subscription_plans的user_id。举个例子，如果一个user_id 4只有一个subscription_plan_id，他就不应该出现。如何只选择那些有多个记录的用户？

如果user_id 5的subscription_plan_id是2和3，他应该显示出来。有关如何做到这一点的任何建议？

select user_id, subscription_plan_id, created_at from 
subscriptions where 
(subscription_plan_id = 47 or 
subscription_plan_id = 78) 
and created_at between '2014-01-01' and '2014-01-31';

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2014-05-02 sharataka

调查这些SQL关键字：

GROUP BY
HAVING
COUNT（DISTINCT [专栏]）

好，好，我会为你做它.... [叽] ...

SELECT 
    user_id, COUNT(DISTINCT subscription_plan_id), MIN(created_at), MAX(created_at) 
FROM 
    subscriptions 
WHERE 
    created_at BETWEEN '2014-01-01' AND '2014-01-31' 
GROUP BY 
    user_id 
HAVING 
    COUNT(DISTINCT subscription_plan_id) > 1 
;

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2014-05-02 15:58:35

“叽” :)。这有效，但只返回第一次订阅的user_id。我如何获得该用户的所有subscription_plan_ids？ – sharataka

我的意思是返回订阅数量。我试图得到实际的subscription_plan_ids。 – sharataka

您需要使用GROUP BY和HAVING筛选出具有订阅一定数量的用户，你的情况不到1 尝试是这样的

select user_id, subscription_plan_id, created_at 
from subscriptions 
where (subscription_plan_id = 47 or 
    subscription_plan_id = 78) 
    and created_at between '2014-01-01' and '2014-01-31' 
    AND user_id IN (SELECT user_id FROM subscriptions ss 
        GROUP BY ss.user_id 
        HAVING COUNT(subscription_plan_id) > 1) ;

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2014-05-02 15:54:41 user2989408

如何只选择那些有多个记录的用户？

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