MySQL多选问题

MySQL多选问题,mysql,Mysql,如何将下面的代码示例1添加到示例2中,而不会弄乱查询 例1 SELECT * FROM users INNER JOIN users_articles ON users.user_id = users_articles.user_id WHERE users.active IS NULL AND users.deletion = 0 例2 SELECT users.user_id, users_articles.user_id, users_articles.title, articles_

如何将下面的代码示例1添加到示例2中,而不会弄乱查询

例1

SELECT *
FROM users 
INNER JOIN users_articles ON users.user_id = users_articles.user_id
WHERE users.active IS NULL
AND users.deletion = 0
例2

SELECT users.user_id, users_articles.user_id, users_articles.title, articles_comments.article_id, articles_comments.comment, articles_comments.comment_id
FROM users_articles
INNER JOIN articles_comments ON users_articles.id = articles_comments.article_id
INNER JOIN users ON articles_comments.user_id = users.user_id
WHERE users.active IS NULL
AND users.deletion = 0
ORDER BY articles_comments.date_created DESC
LIMIT 50

我不确定你到底在问什么,但这有帮助吗

SELECT users.user_id, users_articles.user_id, users_articles.title, articles_comments.article_id, articles_comments.comment, articles_comments.comment_id
FROM users_articles
  INNER JOIN articles_comments ON users_articles.id = articles_comments.article_id
  INNER JOIN users ON articles_comments.user_id = users.user_id
WHERE users.active IS NULL
  AND users.deletion = 0
ORDER BY articles_comments.date_created DESC
LIMIT 50
更新

这是你想要的吗

SELECT users.user_id, users_articles.user_id, users_articles.title, articles_comments.article_id, articles_comments.comment, articles_comments.comment_id
FROM users_articles
  INNER JOIN articles_comments ON users_articles.id = articles_comments.article_id
  INNER JOIN users ON articles_comments.user_id = users.user_id
    AND users.active IS NULL
    AND users.deletion = 0
ORDER BY articles_comments.date_created DESC
LIMIT 50;
SELECT *
FROM users 
  INNER JOIN users_articles ON users.user_id = users_articles.user_id
WHERE users.active IS NULL
  AND users.deletion = 0

在不打乱我的问题的情况下,我做了什么。意思是?我只是不想让我的查询显示错误:@Mitch Wheat我只是想知道如何将示例1添加到示例2中,两个示例中的任何一个都会给您一个错误吗?我不明白你想干什么。在示例1中,除了返回联接中的所有列之外,没有任何一项是示例2没有做过的。@Andrew Cooper我想我想知道如何在示例2中运行另一个select?不,它与示例2完全相同。确切地说,您希望组合查询执行吗?我只想知道如何将示例1中的第一个select添加到示例2中,换句话说,如何运行另一个select?