这是我的数据库:

此查询选择所有补充物:
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这是来自一个子类别的所有补充物(在本例中,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相同补充剂的贝叶斯评分在每个查询上应该是不同的,但在这两个查询上返回的结果是相同的。
下面是我计算第一个查询的贝叶斯评分的部分:
(((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关于第二个问题:
(((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由于某些原因,连接不会对结果产生任何影响。
发布于 2012-11-25 01:31:21
因为评论是补充剂的孩子,加入补充剂不会为评论返回任何不同的东西。
我认为这两个查询返回的结果应该是相同的。
此外,根据AVG的数学定义,术语
(SELECT COUNT(*) FROM reviews) * (SELECT AVG(rating) FROM reviews)等同于
(SELECT SUM(rating) FROM reviews)因此,您可以简化(并加快)您的查询,但将其替换为。
https://stackoverflow.com/questions/13543548
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