根据两个表上的值获取不同的数据集

Question

SELECT users.firstname,
users.lastname,
users.username,
users.image_url,
posts.status,
posts.created_at as created_at,
posts.updated_at as updated_at,
share.post_user_id as user_id,
posts.id,
share.message as message,
share.user_id as share_user_id
FROM `share` INNER JOIN `posts` ON share.post_id = posts.id 
INNER JOIN `users` ON users.id = share.user_id

因此，正如您在此处看到的那样，我正在基于users.id = share.user_id从表users中获取名字，姓氏和用户名。共享表如下：

分享

• 消息

• user_id

• post_user_id

• post_id

现在如何获取基于users.id = share.post_user_id的名称集？

Answer 1

您只需要使用别名（例如user2）在表用户上进行另一个内部联接

    SELECT users.firstname,
    users.lastname,
    users.username,
    users.image_url,
    posts.status,
    posts.created_at as created_at,
    posts.updated_at as updated_at,
    share.post_user_id as user_id,
    posts.id,
    share.message as message,
    share.user_id as share_user_id,
    users2.firstname,
    users2.lastname,
    users2.username,
    FROM `share` 
    INNER JOIN `posts` ON share.post_id = posts.id 
    INNER JOIN `users` ON users.id = share.user_id
    INNER JOIN `users`  AS user2 ON users2.id = post_user_id.user_id