mysql - 如何过滤 MySQL 结果以在每个用户的基础上显示 LIMIT 1？

Question

这是 SQL 命令结果的屏幕截图:

这是SQL命令:

SELECT inl_cbsubs_subscriptions.user_id AS cbsubsUserID, inl_cbsubs_subscriptions.id AS cbsubsId, inl_cbsubs_subscriptions.status AS status, inl_cbsubs_payment_items.subscription_id     AS paymentSubID, inl_cbsubs_payment_items.stop_date AS paymentStopDate, inl_cbsubs_payment_items.id AS paymentID
FROM inl_cbsubs_subscriptions
INNER JOIN inl_cbsubs_payment_items
ON inl_cbsubs_subscriptions.id=inl_cbsubs_payment_items.subscription_id
WHERE status='C'
ORDER BY paymentID DESC;

我希望调整此命令，以便我只显示每个用户的最新结果。所以换句话说，在这种情况下，表格应该类似于:

如您所见，cbsubsUserID 每个 ID 仅显示一个结果，而之前 596 id 有多个结果。

Answer 1

如果你想要最新的结果，最好的方法是使用not exists:

SELECT s.user_id AS cbsubsUserID, s.id AS cbsubsId, s.status AS status,
       i.subscription_id AS paymentSubID, i.stop_date AS paymentStopDate, i.id AS paymentID
FROM inl_cbsubs_subscriptions s INNER JOIN
     inl_cbsubs_payment_items i
     ON s.id = i.subscription_id
WHERE s.status = 'C' and
      not exists (select 1
                  from inl_cbsubs_payment_items i2
                  where i2.subscription_id = i.subscription_id and
                    i2.id > i.id
                 )
ORDER BY paymentID DESC;

您需要在 inl_cbsubs_payment_items(subscription_id, paymentid) 上建立索引。

这句话的意思是:“为我提供给定订阅 ID 的所有项目，这些项目没有更大的订阅付款停止日期”。这是一种奇特的说法，“给我每个订阅 ID 的最新信息”，但它往往在数据库中最有效。