Mysql 如何在单个查询中进行选择？

我试图从mysql中的4个表中选择两个（逗号分隔）项，一个用于“菜系”，另一个用于“设施”

我就是这样尝试的：

$query = "SELECT GROUP_CONCAT(DISTINCT c.name SEPARATOR ', ') AS cuisines,
                 GROUP_CONCAT(DISTINCT f.name SEPARATOR ', ') AS facilities
          FROM cuisines c, facilities f
            INNER JOIN restaurant_cuisines rc ON rc.cuisine_id = c.id   
            INNER JOIN restaurant_facilities rf ON rf.facility_id = f.id    
          WHERE rc.restaurant_id = 16 AND rf.restaurant_id = 16"; 

但这个查询不起作用。但是，如果我对这两个选择使用两个单独的查询，那么我可以让它工作。但我正在寻找一个单一查询的解决方案

如果我使用两个单独的查询，这就是输出：

mysql> SELECT GROUP_CONCAT(DISTINCT c.name SEPARATOR ', ') AS cuisines
    -> FROM cuisines c
    -> INNER JOIN restaurant_cuisines rc ON rc.cuisine_id = c.id
    -> WHERE rc.restaurant_id = 15;
+-----------------------------------------+
| cuisines                                |
+-----------------------------------------+
| American, Asian, Bars & Pubs, Beverages |
+-----------------------------------------+
1 row in set (0.14 sec)

有人能告诉我mysql中有可能吗

---

多谢各位

要使GROUP BY语句生效，您需要添加GROUP BY语句：

SELECT cuisines,facilities FROM (
SELECT GROUP_CONCAT(DISTINCT c.name SEPARATOR ', ') AS cuisines, restaurant_id
      FROM cuisines c
        INNER JOIN restaurant_cuisines rc ON rc.cuisine_id = c.id   
      WHERE rc.restaurant_id = 16) a
INNER JOIN (
SELECT GROUP_CONCAT(DISTINCT f.name SEPARATOR ', ') AS facilities,restaurant_id
      FROM facilities f
        INNER JOIN restaurant_facilities rf ON rf.facility_id = f.id    
      WHERE rf.restaurant_id = 16
      GROUP BY rf.restaurant_id) b ON a.restaurant_id=b.restaurant_id;

---

不要在from子句中混合使用

、

和

内部联接

符号。它们是两个独立的标准（ANSI和非ANSI）。编译器必须选择并使用其中一个，不能同时使用这两个标准。换句话说

FROM cuisines c, <-- This isn't allowed with inner join syntax.. 
     facilities f
INNER JOIN restaurant_cuisines rc ON rc.cuisine_id = c.id   
INNER JOIN restaurant_facilities rf ON rf.facility_id = f.id   

或

然而，我相信你错过了餐厅烹饪和餐厅设施之间的一个连接点，可能是在餐厅Id上，这将允许你消除这两个

rc.restaurant\u Id=16
和rf.restaurant_id=16

请使用您想要的输出（以及两个表中餐馆16的输入）进行更新@amdixon，更新了我的问题。是

restaurant\u id

对于这两个表都是相同的。您的查询是混合、表示法和内部联接表示法。这就是为什么会出现错误选择一个。错误1054（42S22）：“字段列表”中的未知列“设施”抱歉，没有完成查询。请参阅1分钟后编辑错误1054（42S22）：字段列表中的未知列“c.name”重试：）复制/粘贴发布错误1054（42S22）：on子句中的未知列“a.restaurant_id”

SELECT GROUP_CONCAT(DISTINCT c.name SEPARATOR ', ') AS cuisines,
       GROUP_CONCAT(DISTINCT f.name SEPARATOR ', ') AS facilities
FROM cuisines c
INNER JOIN restaurant_cuisines rc 
  ON rc.cuisine_id = c.id   
INNER JOIN restaurant_facilities rf 
  ON rc.restraunt_Id = Rf.restraunt_ID
INNER JOIN facilities f
  ON rf.facility_id = f.id    
WHERE rc.restaurant_id = 16; 

SELECT GROUP_CONCAT(DISTINCT c.name SEPARATOR ', ') AS cuisines,
       GROUP_CONCAT(DISTINCT f.name SEPARATOR ', ') AS facilities
FROM cuisines c
    ,facilities f
    , restaurant_cuisines rc 
    , restaurant_facilities rf  
WHERE rc.restaurant_id = 16 
  AND rf.restaurant_id = 16
  AND rc.cuisine_id = c.id   
  AND rf.facility_id = f.id