SQL查詢選擇相同的數據

Question

在我的查詢中，rating_food,rating_service和rating_decor對於我的所有結果都返回相同的值。

select `restaurants`.*, `rating`.*, (rating.rating_food + rating.rating_service + rating.rating_decor)/3 as rating_total from 
(
SELECT 
avg(reviews.rating_food) * (- 1/(0.33 * count(reviews.id) + 1) + 1) + 5 * 1/(count(reviews.id) + 1) as rating_food, 
avg(reviews.rating_service) * (- 1/(0.33 * count(reviews.id) + 1) + 1) + 5 * 1/(count(reviews.id) + 1) as rating_service, 
avg(reviews.rating_decor) * (- 1/(0.33 * count(reviews.id) + 1) + 1) + 5 * 1/(count(reviews.id) + 1) as rating_decor 
FROM restaurants 
JOIN reviews ON reviews.restaurant_id = restaurants.id 
) 
as rating, restaurants 
order by `rating_total` asc limit 12 

預期結果：
12個結果的陣列與另外三個計算列（rating_food等）

現在，我得到12倍的結果，但他們所有的rating_food，rating_service和rating_decor返回相同的值的所有結果。

Answer 1

您需要將ratings子查詢與restaurants表連接起來。

試試這個：

select `restaurants`.*, `rating`.*, 
(rating.rating_food + rating.rating_service + rating.rating_decor)/3 as rating_total from 
(
SELECT 
restaurants.id, 
avg(reviews.rating_food) * (- 1/(0.33 * count(reviews.id) + 1) + 1) + 5 * 1/(count(reviews.id) + 1) as rating_food, 
avg(reviews.rating_service) * (- 1/(0.33 * count(reviews.id) + 1) + 1) + 5 * 1/(count(reviews.id) + 1) as rating_service, 
avg(reviews.rating_decor) * (- 1/(0.33 * count(reviews.id) + 1) + 1) + 5 * 1/(count(reviews.id) + 1) as rating_decor 
FROM restaurants 
JOIN reviews ON reviews.restaurant_id = restaurants.id 
GROUP by restaurants.id 
) 
as rating join restaurants on rating.id = restaurants.id 
order by `rating_total` asc limit 12 

SQL小提琴：http://sqlfiddle.com/#!9/50b39/4/0