CASE和INNER JOIN的MySQL查询错误

Question

我有一个表的用户信息被称为'a'和一个表来自不同API（twitter，foursquare）的数据，在这个例子中，基于a.api_type的值 应该成为b。我最终想要的是能够从活动API（api_foursquare或api_twitter）中获取正确的头像。

我一直在试图让这个使用这个查询一段时间，但我不断收到此错误。 SQL是不是我的最强点，那么如何解决这个问题的任何提示将是巨大的:)

"[Err] 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 

    INNER JOIN b ON b.user_id = a.user_id at line 11" 

SELECT a.user_id, 
     a.api_type, 
     b.avatar, 
     b.user_id 
(CASE 
     WHEN a.api_type = 0 THEN api_foursquare 
     WHEN a.api_type = 1 THEN api_twitter 
END) as b 
FROM a WHERE a.cookie_hash = :cookie_hash 
INNER JOIN b ON b.user_id = a.user_id 

Answer 1

你不能选择一个表加入一个case语句。你应该离开联接两个表，然后从他们中挑选一个值在case声明，就像这样：

SELECT a.user_id, 
     a.api_type, 
     (case WHEN a.api_type = 0 THEN b.avatar ELSE c.avatar END) as avatar, 
     (case WHEN a.api_type = 0 THEN b.user_id ELSE c.user_id END) as user_id 
FROM a 
    LEFT OUTER JOIN api_foursquare b ON b.user_id = a.user_id 
    LEFT OUTER JOIN api_twitter c ON c.user_id = a.user_id 
WHERE a.cookie_hash = :cookie_hash 

你可能并不需要最后的表达式（...as user_id之一），因为它将等于a.user_id，如果api_twitter或api_foursquare中有一行匹配a.user_id。

你也必须把WHERE子句FROM条款后：

EDIT：考虑到ypercube的伟大的建议，查询应该是这样的：

SELECT a.user_id, 
     a.api_type, 
     COALESCE(b.avatar, c.avatar) as avatar 
FROM a 
    LEFT OUTER JOIN api_foursquare b ON b.user_id = a.user_id AND a.api_type = 0 
    LEFT OUTER JOIN api_twitter c ON c.user_id = a.user_id and a.api_type = 1 
WHERE a.cookie_hash = :cookie_hash 

Answer 2

您b.user_id后(CASE之前需要逗号：

SELECT a.user_id, 
    a.api_type, 
    b.avatar, 
    b.user_id, 
    (CASE 
      WHEN a.api_type = 0 THEN api_foursquare 
      WHEN a.api_type = 1 THEN api_twitter 
    END) as b 
FROM a WHERE a.cookie_hash = :cookie_hash 
    INNER JOIN b ON b.user_id = a.user_id