Question

在我的查询中，rating_food，rating_service和rating_decor会为我的所有结果返回相同的值。

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

预期结果：
一个包含12个结果的数组，其中包含三个额外的计算列（rating_food等）

现在我得到了12个结果，但他们的所有rating_food，rating_service和rating_decor都返回了所有结果的相同值。

Answer 1

您需要将nodeName子查询与ratings表联接回来。

试试这个：

restaurants

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

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