Question

使用MySQL我试图提取所有people company_id匹配3的人person，并比较favourites表中是否存在job_id。

我注意到，如果一个人没有people，那么他们就不会通过以下查询输出。

以下查询输出1行，但在company_id表中有3行与3的{{1}}匹配

                SELECT *,
                           j.company_id as companyid,
                           f.id IS NOT NULL AS jid,
                           p.id as pid,
                           f.id AS fave_id,
                           f.id IS NOT NULL AS fave
                      FROM people p  
                INNER JOIN job j 
                        ON p.job_id = j.id
                 LEFT JOIN favourites f
                        ON f.people_id=p.id
                       AND f.user_id = 12
                     WHERE p.company_id = 3
                  ORDER BY p.id ASC

SQLFiddle

感谢任何帮助，让查询输出所有匹配的行，并检查favourites表。

Answer 1

如果你把左联接放在工作上，它会显示没有job_id的人。我更新了小提琴here。

SELECT
    p.*,
    j.company_id as companyid,        
    f.id IS NOT NULL AS jid,
    p.id as pid,
    f.id AS fave_id,
    f.id IS NOT NULL AS fave
FROM people p  
LEFT JOIN job j 
       ON j.id = p.job_id
LEFT JOIN favourites f 
       ON f.people_id=p.id
      AND f.user_id = 12
WHERE p.company_id = 3
ORDER BY p.id ASC

Answer 2

您应该将作业表的INNER JOIN更改为LEFT OUTER JOIN，以便无论它们是否存在于作业表中，都可以选择它们。

MySQL从多行中选择

2 个答案: