从多对多表中添加总计和存在聚合的高效 SQL 查询答案

【问题标题】：Efficient SQL query to add totals and existence aggregates from a many-many table从多对多表中添加总计和存在聚合的高效 SQL 查询
【发布时间】：2019-07-15 23:56:21
【问题描述】：

我有两张桌子

杂志与字段id, name;
Subscriptions 包含字段 subscriber_id 和 magazine_id，它们是来自 Magazines 和另一个订阅者表的外键。

我想将 Magazines 表扩展为：

第三列列出其订阅者和
第四列 subscribed 有一个布尔值，如果 Subscriptions 中存在一条记录，且我的subscriber_id 为真（我们可以说我是 subscriber_id=1 ）。

是否有解决这些问题的规范方法？如果没有，我怎样才能有效地做到这一点？

要获得订阅者列，我想到的最简单的方法是不列出没有订阅者的杂志。（我确实读过here，但看不出它是如何相关的）

SELECT magazine_id, COUNT(*)
FROM subscriptions GROUP BY magazine_id
ORDER BY magazine_id ASC;

所以我做了一些有点老套而且可能效率低下的事情：

SELECT
 magazines.id,
 CASE WHEN NOT EXISTS(
  SELECT * FROM subscriptions WHERE magazines.id=magazine_id
 ) THEN 0
 ELSE COUNT(*) END AS subscribers
FROM
 magazines
LEFT JOIN
  subscriptions
ON
 magazines.id = subscriptions.magazine_id
GROUP BY
 magazines.id
ORDER BY
 magazines.id ASC;

对于布尔型第四列，我得到了一个解决方案，方法是再次将先前的结果与订阅一起左连接，并为subscriber_id 的值做 case when is null。

我正在使用 PostgreSQL。我不是学生，但我想学习如何解决这个问题，杂志/订阅者只是一个例子。我只知道 SQL 的基础知识。

【问题讨论】：

标签： sql aggregate-functions rdbms query-performance sqlperformance

【解决方案1】：

第一件事是获取订阅者数量的查询并将其加入杂志表。

所以

SELECT id, COUNT(*) AS subscriber_count
FROM subscriptions GROUP BY magazine_id

变成

SELECT mag.id, COALESCE(subs.subscriber_count, 0) AS subscribers
FROM magazines mag
LEFT JOIN (
    SELECT magazine_id, COUNT(*) AS subscriber_count 
    FROM subscriptions GROUP BY magazine_id) subs ON subs.magazine.id = mag.id
ORDER BY mag.id ASC

如果您是订阅者，则可以通过以下几种方式进行添加： 1）与上面类似，创建一个查询并将其加入杂志

--Query:
SELECT DISTINCT magazine_id FROM subscriptions WHERE subscriber_id = 1

--Joined:
SELECT 
    mag.id, 
    COALESCE(subs.subscriber_count, 0) AS subscribers,
    CASE WHEN mysubs.magazine_id IS NULL THEN 0 ELSE 1 END AS my_subscription
FROM magazines mag
LEFT JOIN (
    SELECT magazine_id, COUNT(*) AS subscriber_count 
    FROM subscriptions GROUP BY magazine_id) subs ON subs.magazine_id = mag.id
LEFT JOIN (
    SELECT DISTINCT magazine_id FROM subscriptions WHERE subscriber_id = 1
    ) mysubs ON mysubs.magazine_id = mag.id
ORDER BY mag.id ASC

2) 如果订阅表在杂志 ID 和订阅者 ID 上是唯一的，您可以直接加入而无需子查询：

SELECT 
    mag.id, 
    COALESCE(subs.subscriber_count, 0) AS subscribers,
    CASE WHEN mysubs.magazine_id IS NULL THEN 0 ELSE 1 END AS my_subscription
FROM magazines mag
LEFT JOIN (
    SELECT magazine_id, COUNT(*) AS subscriber_count 
    FROM subscriptions GROUP BY magazine_id) subs ON subs.magazine_id = mag.id
LEFT JOIN subscriptions mysubs on mysubs.magazine_id = mag.id AND subscriber_id = 1
ORDER BY mag.id ASC

3) 使用 bool 上的 MAX 聚合将条件添加到订阅子查询

SELECT 
    mag.id, 
    COALESCE(subs.subscriber_count, 0) AS subscribers,
    COALESCE(subs.mysub, 0) AS my_subscription
FROM magazines mag
LEFT JOIN (
    SELECT magazine_id, COUNT(*) AS subscriber_count,
        MAX(CASE WHEN subscriber_id = 1 THEN 1 ELSE 0 END) AS mysub
    FROM subscriptions GROUP BY magazine_id) subs ON subs.magazine_id = mag.id
ORDER BY mag.id ASC

【讨论】：

【解决方案2】：

如果你只想要杂志ID，你可以这样做：

select magazine_id, count(*) as num_subscribers,
       max(case when subscriber_id = 1 then 1 else 0 end) as my_subscriber
from subscriptions s
group by magazine_id;

也就是说，如果您只需要杂志 ID，则不需要 join。

注意：这只会返回有订阅者的杂志。

这可以通过left join 轻松扩展：

select m.mag, count(s.magazine_id) as num_subscribers,
       max(case when s.subscriber_id = 1 then 1 else 0 end) as my_subscriber
from magazines m left join
     subscriptions s
     on s.magazine_id = m.mag_id
group by m.mag_id;

【讨论】：