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これが私のデータベースです：

ここに画像の説明を入力してください

このクエリはすべてのサプリメントを選択します：

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

そしてこれは、1つのサブカテゴリ（この場合、id = 1）からのすべてのサプリメントです。

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

同じサプリメントのベイジアン評価は、クエリごとに異なるはずですが、両方で同じものを返します。

これが、最初のクエリでベイジアン評価を計算する部分です。

(((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

2番目に：

(((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

何らかの理由で、結合によって結果に違いが生じることはありません。

4

1 に答える 1

1

レビューはサプリメントの子であるため、サプリメントに参加しても、レビューに異なるものは返されません。

2つのクエリが同じものを返すことが期待されていると思います。


また、の数学的定義によりAVG、用語

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

と同じです

(SELECT SUM(rating) FROM reviews)

したがって、クエリを単純化（および高速化）できますが、これをで置き換えます。

于 2012-11-24T17:31:21.263 に答える