MySQL 获取其余值

Question

使用此查询

SELECT DISTINCT(job_primary.id) AS id 
FROM `job_primary` 
LEFT JOIN `job_skill` ON `job_skill`.`job_id` = `job_primary`.`id` 
LEFT JOIN `job_facts` ON `job_facts`.`job_id` = `job_primary`.`id`  
LEFT JOIN `job_location` ON `job_location`.`job_id` = `job_primary`.`id`  
LEFT JOIN `job_people` ON `job_people`.`job_id` = `job_primary`.`id`   
LEFT JOIN `job_sec` ON `job_sec`.`job_id` = `job_primary`.`id` 
                    AND job_sec.exp_start_date > UNIX_TIMESTAMP() 
INNER JOIN `user_skills` ON `user_skills`.`skill_id` = `job_skill`.`skill_id` 
                         AND user_skills.user_id = 1 
WHERE job_primary.posted_by != 1

我只得到 id 值。但我想获得具有不同 id 的其余值。不要考虑其余的代码。问题就在这里，如果我使用:

select *, job_primary.id as id

选择代码然后我获取所有数据。但它并不明显。所以如果我使用

SELECT *, distinct(job_primary.id) as id

或

SELECT distinct(job_primary.id), * as id

但它显示错误。

Answer 1

我认为您的困惑在于查询本身的内容，nixon1333。您想向数据库询问什么？如果您只想每个 id 一行，请使用 SELECT * FROM ... GROUP BY job_primary.id

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