php - 子选择查询中的表名别名范围

Question

请看下面的查询 - 我得到Unknown column 'u.id' in 'on clause'

SELECT id, username, 

coalesce(
 (SELECT name from company c 
    INNER JOIN user_company uc 
    ON uc.user_id = u.id 
    AND c.id = uc.company_id), 'NOT-AVAILABLE' 
) companyname

FROM `user` u

Answer 1

根据您的评论，相关性不能放在JOIN相关子查询的范围内。

但是，它可以放在WHERE相关子查询的子句中。

SELECT
  id,
  username,
  coalesce(
    (SELECT name
       FROM company      c
 INNER JOIN user_company uc
         ON c.id = uc.company_id
      WHERE uc.user_id = u.id
    ),
    'NOT-AVAILABLE'
  ) companyname
FROM
  `user` u 

这回答了你的明确问题；为什么您的查询在语法上失败。

但是，我会用一个简单的LEFT JOIN.

SELECT
  u.id,
  u.username,
  COALESCE(c.name, 'NOT-AVAILABLE') companyname
FROM
  `user` u 
LEFT JOIN
  `user_company` uc
      ON uc.user_id = u.id
LEFT JOIN
  `company` c
    ON c.id = uc.company_id

Answer 2

一种解决方法：

SELECT u.id,
       u.username,
       COALESCE(aux.name, 'NOT-AVAILABLE') as 'companyname'
FROM `user` u
LEFT JOIN
 (SELECT user_id, name from company c 
    INNER JOIN user_company uc
    ON c.id = uc.company_id) aux ON aux.user_id = u.id

---

另一种解决方法：

SELECT u.id,
       u.username,
       COALESCE(c.name, 'NOT-AVAILABLE') AS 'companyname'
FROM `user` u
LEFT JOIN user_company uc ON uc.user_id = u.id 
LEFT JOIN company c ON c.id = uc.company_id