我有 2 个表：
餐厅
+---------------+----------------+
| restaurant_id | Restaurant name|
+---------------+----------------+
| 1 | KFC |
| 2 | McD |
+----+---------------------------+
菜
+---------------+---------------+----------------+
| cuisine_id | restaurant_id | cuisine |
+---------------+---------------+----------------+
| 1 | 1 | Fastfood |
| 2 | 1 | Fried Chicken |
| 3 | 2 | Fastfood |
| 4 | 2 | Burger |
+---------------+---------------+----------------+
是否有可能获得这样的数据=
+---------------+----------------+------------------------+
| restaurant_id | Restaurant name| Cuisine |
+---------------+----------------+------------------------+
| 1 | KFC |Fastfood, Fried Chicken |
| 2 | McD |Fastfood, Burger |
+----+---------------------------+------------------------+
使用单个查询？或者我应该通过 php 来做（首先选择 table Restaurant> foreach 循环>按餐厅 id 选择美食>解析到新数组中）。
您可以使用GROUP_CONCAT来实现此目的：
SELECT r.restaurant_id, r.Restaurant_name,
GROUP_CONCAT(c.cuisine ORDER BY c.cuisine)
FROM Restaurant AS r
LEFT JOIN Cuisine AS c ON r.restaurant_id = c.restaurant_id
GROUP BY r.restaurant_id, r.Restaurant_name
注意：仅当您想要一个有序的逗号分隔的cuisine名称列表时，才需要GROUP_CONCAT中的ORDER BY子句。
您可以使用
GROUP_CONCAT：
SELECT c.restaurant_id,r.restaurant_name, GROUP_CONCAT(c.cuisine)
FROM Restaurant r
INNER JOIN Cuisine c ON c.restaurant_id = r.restaurant_id
GROUP BY c.restaurant_id
SELECT Restaurant.restaurant_id, Restaurant.Restaurant_name, GROUP_CONCAT(Cuisine.cuisine)
FROM Restaurant
LEFT JOIN Cuisine ON Restaurant.restaurant_id = Cuisine.restaurant_id
GROUP BY Restaurant.restaurant_id, Restaurant.Restaurant_name