mysql - 多个 LEFT JOIN - 输出错误

Question

我收到这个查询:

SELECT r.name, date(r.join_date) as join_date,
(COUNT(DISTINCT a.id) + COUNT(DISTINCT n.id)) as num_likes,
COALESCE(SUM(rv.views), 0) as sum_views
FROM restaurants r
LEFT JOIN apple_likes a ON r.id = a.restaurant_id
LEFT JOIN android_likes n ON r.id = n.restaurant_id
LEFT JOIN restaurant_views rv on r.id = rv.restaurant_id
WHERE r.id=192
GROUP BY r.id DESC

这是另一个查询，其中删除了 num_like 变量和 LEFT JOIN:

SELECT r.name, date(r.join_date) as join_date,
COALESCE(SUM(rv.views), 0) as sum_views
FROM restaurants r
LEFT JOIN restaurant_views rv on r.id = rv.restaurant_id
WHERE r.id=192
GROUP BY r.id DESC

如您所见，sum_views 更改为 1793，这是正确的!
意思是，当我删除其他 LEFT JOIN 时，sum_views 显示正确的值。
如何使上层查询在 sum_views 上显示正确的值？
(如果需要，我可以上传其他表格的结构图像!)

<小时/>

编辑:
我找到了一种使用子查询来解决问题的方法

SELECT r.name, date(r.join_date) as join_date,
(COUNT(DISTINCT a.id) + COUNT(DISTINCT n.id)) as num_likes,
(SELECT COALESCE(SUM(views), 0) FROM restaurant_views WHERE restaurant_id=192) as sum_views
FROM restaurants r
LEFT JOIN apple_likes a ON r.id = a.restaurant_id
LEFT JOIN android_likes n ON r.id = n.restaurant_id
WHERE r.id=192
GROUP BY r.id DESC

如果有人知道如何在不使用子查询的情况下解决这个问题，我很有兴趣听到它!

Answer 1

您的连接正在乘以行数。一种解决方案是在连接之前预先聚合数据:

SELECT r.name, date(r.join_date) as join_date,
       (coalesce(a.cnt, 0) + coalesce(n.cnt, 0)) as num_likes,
       COALESCE(SUM(rv.views), 0) as sum_views
FROM restaurants r LEFT JOIN
     (select a.restaurant_id, count(*) as cnt
      from apple_likes a
      group by a.restaurant_id
     ) a
     ON r.id = a.restaurant_id LEFT JOIN
     (select n.restuarant_id, count(*) as cnt
      from android_likes n
      group by n.restaurant_id
     ) n
     ON r.id = n.restaurant_id LEFT JOIN
     (select rv.restaurant_id, count(*) as cnt
      from restaurant_views rv
      group by rv.restaurant_id
     ) rv
     on r.id = rv.restaurant_id
WHERE r.id = 192
GROUP BY r.id DESC