MySQL加入不在子查询上

Question

这是我的数据库：

此查询选择所有补充：

SELECT a.id, a.name, a.image, a.url_segment, COUNT(b.id) AS reviews_count, ROUND(AVG(b.rating), 2) AS reviews_rating, (((SELECT COUNT(*) FROM reviews) * (SELECT AVG(rating) FROM reviews)) + (COUNT(b.id) * AVG(b.rating))) / ((SELECT COUNT(*) FROM reviews) + COUNT(b.id)) AS bayesian_rating
FROM (`supplements` AS a)
LEFT JOIN `reviews` AS b ON `b`.`supplements_id` = `a`.`id`
GROUP BY `a`.`id`
ORDER BY `bayesian_rating` DESC

这是所有补充来自一个子类别（在这种情况下，id = 1）：

SELECT a.id, a.name, a.image, a.url_segment, COUNT(b.id) AS reviews_count, ROUND(AVG(b.rating), 2) AS reviews_rating, (SELECT text FROM reviews WHERE supplements_id = a.id ORDER BY id DESC LIMIT 1) AS reviews_latest_text, (((SELECT COUNT(*) FROM reviews LEFT JOIN supplements ON (supplements.id = reviews.supplements_id AND supplements.subcategories_id = 1)) * (SELECT AVG(rating) FROM reviews LEFT JOIN supplements ON (supplements.id = reviews.supplements_id AND supplements.subcategories_id = 1))) + (COUNT(b.id) * AVG(b.rating))) / ((SELECT COUNT(*) FROM reviews LEFT JOIN supplements ON (supplements.id = reviews.supplements_id AND supplements.subcategories_id = 1)) + COUNT(b.id)) AS bayesian_rating
FROM (`supplements` AS a)
LEFT JOIN `reviews` AS b ON `b`.`supplements_id` = `a`.`id`
WHERE `a`.`subcategories_id` =  '1'
GROUP BY `a`.`id`
ORDER BY `bayesian_rating` DESC

相同补充的贝叶斯评级在每个查询中应该是不同的，但两者都返回相同。

这是我在第一个查询中计算贝叶斯评级的部分：

(((SELECT COUNT(*) FROM reviews) * (SELECT AVG(rating) FROM reviews)) + (COUNT(b.id) * AVG(b.rating))) / ((SELECT COUNT(*) FROM reviews) + COUNT(b.id)) AS bayesian_rating

关于第二个：

(((SELECT COUNT(*) FROM reviews LEFT JOIN supplements ON (supplements.id = reviews.supplements_id AND supplements.subcategories_id = 1)) * (SELECT AVG(rating) FROM reviews LEFT JOIN supplements ON (supplements.id = reviews.supplements_id AND supplements.subcategories_id = 1))) + (COUNT(b.id) * AVG(b.rating))) / ((SELECT COUNT(*) FROM reviews LEFT JOIN supplements ON (supplements.id = reviews.supplements_id AND supplements.subcategories_id = 1)) + COUNT(b.id)) AS bayesian_rating

由于某种原因，加入对结果没有任何影响。

Answer 1

由于评论是补充剂的孩子，加入补充剂不会返回任何不同的评论。

我认为期望两个查询返回相同的内容。

此外，通过AVG的数学定义，术语

(SELECT COUNT(*) FROM reviews) * (SELECT AVG(rating) FROM reviews)

与

相同

(SELECT SUM(rating) FROM reviews)

因此您可以简化（并加快）您的查询，但在此代替。