Question

我有以下表格：

作者：

id username email password salt email_salt email_verified IP_ADDRESS

Author_threads：

thread_id，author_id

主题：

id，title，content，created

标签：

id，name

Thread_tags：

tad_id，thread_id

我想选择最新的30个主题，他们的作者及其所有标签。这是我使用的SQL语句：

       SELECT thread.title, thread.id as thread_id,
       thread.content, author.username, author.id as author_id,
       GROUP_CONCAT(DISTINCT tag.name ORDER BY tag.name DESC SEPARATOR ',') AS tags
       FROM thread 
       JOIN thread_tags ON thread.id = thread_tags.thread_id
       JOIN tag ON thread_tags.tag_id = tag.id

       JOIN author_threads ON thread.id = author_threads.thread_id
       JOIN author ON author_threads.author_id = author.id

       GROUP BY thread.id DESC
       LIMIT 0, 30

有更简单的方法吗？

Answer 1

我认为GROUP BY应该是ORDER BY？

除此之外，我确实相信你这样做是正确的。

当您需要所有表中的值时，您需要将它们全部加入，这就是您所做的。

Answer 2

您唯一的另一个选择是使用多表选择，例如

   SELECT thread.title, thread.id as thread_id,
   thread.content, author.username, author.id as author_id,
   GROUP_CONCAT(DISTINCT tag.name ORDER BY tag.name DESC SEPARATOR ',') AS tags
   FROM thread, thread_tags, tag, author_threads, author
   WHERE thread.id = thread_tags.thread_id
       AND thread_tags.tag_id = tag.id
       AND thread.id = author_threads.thread_id
       AND author_threads.author_id = author.id
   GROUP BY thread.id DESC
   LIMIT 0, 30

但对此的支持可能与数据库有关。

更简单的使用连接的方式

2 个答案: