多个LEFT JOIN - 输出错误

Question

我收到了这个问题：

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

以下是另一个使用 num_like 变量并删除LEFT JOIN的查询：

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！
这意味着，当我删除其他LEFT JOIN时，sum_views显示正确的值 如何使上部查询在sum_views上显示正确的值？
（如果需要，我可以上传其他表格结构的图片！）

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修改
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，我很乐意听到它！

Answer 1

您的join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

Answer 2

所以试试：

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
INNER JOIN restaurant_views rv on r.id = rv.restaurant_id
WHERE r.id=192
GROUP BY r.id DESC;

Answer 3

使用此：

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
    JOIN apple_likes a ON r.id = a.restaurant_id
    JOIN android_likes n ON r.id = n.restaurant_id
    JOIN restaurant_views rv on r.id = rv.restaurant_id
  WHERE r.id=192
  GROUP BY r.id DESC;

请注意，我只删除关键字left，因此将其设为natural join，以便在计算所有null joins时不会包含该项。