Mysql SQL查询选择相同的数据

在我的查询中，

rating\u food

、

rating\u service

和

rating\u decor

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个结果和三个额外计算列的数组（

rating\u food

etc）

---

现在我得到了12个结果，但是他们所有的

rating\u food

、

rating\u service

和

rating\u decor

对所有结果返回相同的值。

您需要将

ratings

子查询与

restaurants

表连接起来

试试这个：

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 FIDLE:

是否在派生表中添加GROUPBY子句？和主查询连接条件编辑您的问题并提供示例数据和所需结果。这可能会让你感到惊讶，但一个不起作用的查询并不一定能很好地传达你想要做的事情。@GordonLinoff已经添加了一个解释。为什么评级和餐馆之间没有联接条件？正如前面所写的，你正在交叉联接所有评级的餐馆。您没有关联它们。这将返回1个结果。预期已修复，但缺少一组。还链接了新的SQL Fiddle。这就解决了它！我添加了左连接评论以获取所有结果，即使它们与任何评论都没有关系。