Mysql 在一个表中计算三个可能性

我有一个带有订阅服务器的表，它有两个字段，actived和suscribed。我需要得到总的订阅者数量，激活的数量和在同一咨询中订阅的数量。我曾尝试使用双左连接来实现这一点，但我需要按一个不是主键的字段分组，结果出现错误。我现在有这样的咨询：

FROM (SELECT subscribers.mailing_list, subscribers.mailing_list AS suscriptores, s2.inactivos AS inactivos, s3.excluidos AS excluidos
FROM subscribers
LEFT JOIN (SELECT id, mailing_list, COUNT(*) AS inactivos FROM subscribers WHERE subscribed = false GROUP BY id) s2 ON subscribers.id = s2.id
LEFT JOIN (SELECT id, mailing_list, COUNT(*) AS excluidos FROM subscribers WHERE excluded = true GROUP BY id) s3 ON subscribers.id = s3.id
) AS subs
GROUP BY subs.mailing_list```

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不需要3个查询，而是需要聚合函数中的

case表达式

，这称为“条件聚合”，例如

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在MySQL中，您可以将此逻辑简化为：

SELECT mailing_list,
       SUM( subscribed = FALSE ) AS inactivos,
       SUM( subscribed = TRUE ) AS excluidos,
       COUNT( * ) AS Total
FROM subscribers
GROUP BY mailing_list;

如果

subscribed

实际上是

boolean

，则可以进一步简化为：

SELECT mailing_list,
       SUM( NOT subscribed ) AS inactivos,
       SUM( subscribed ) AS excluidos,
       COUNT( * ) AS Total
FROM subscribers
GROUP BY mailing_list;

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好极了！（但我确实对西班牙标签感到困惑：）是的，这是另一种解决方案，但我更喜欢0个国际标签null@DonProt肖恩。如何从这些中获取

NULL

？唯一的方法是如果订阅的

为
NULL
，这在其他方面似乎有问题。
SELECT mailing_list,
       SUM( NOT subscribed ) AS inactivos,
       SUM( subscribed ) AS excluidos,
       COUNT( * ) AS Total
FROM subscribers
GROUP BY mailing_list;