Question

我正在尝试选择两个（逗号分隔的）项目，一个用于＆＃34;菜肴和＃34;还有一个用于＆＃34;设施＆＃34;来自mysql中的4个表。

这是我尝试的方式：

$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";

但是这个查询不起作用。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。但我正在寻找单一查询的解决方案。

如果我使用两个单独的查询，这是输出：

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)

任何人都可以告诉我在mysql中是否可能？

谢谢。

Answer 1

您需要添加GROUP BY语句才能生效：

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;

Answer 2

不要在from子句中混合,和inner join符号它们是两个独立的标准（ANSI与非ANSI），编译器必须选择并使用一个，它可以＆＃39 ; t两者兼用。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

我改变了连接的顺序，以便美食 - ＆gt; restraunt_Cusines - ＆GT; restraunt_facilities - ＆GT;设施。我认为这是基于符号和连接的准确。我还删除了where子句标准之一，因为它现在基于创建的连接是多余的。

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;

或

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

但是我相信你错过了restaurant_cuisines和restaurant_facility之间的联接，很可能在restaurant_Id上，这可以让你消除两个rc.restaurant_id = 16 AND rf.restaurant_id = 16

如何在单个查询中进行此选择？

2 个答案: