mysql - 仅获取表中存在的行？

我需要从表中选择所有行：Announcements条件下，用户在表Categories_announcements中存在行：
Categories_user.category_id = Categories_announcements.category_id
我尝试了SQL查询，请参见link
它应该只返回一行Announcements.id = 1，因为用户在Categories_user.category_id中只有一个类别。
编辑：
我已经测试了您共享的SQL，因此这是query：

select *
from `announcements`
where exists (
        select 1
        from `announcement_category`
        inner join `user_category` on `user_category`.`category_id` = `announcement_category`.`category_id`
        where `Auser_category`.`user_id` = 1
            and `announcement_category`.`announcement_id` = announcements.id
        )

它返回一个错误：
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ') LIMIT 0, 25' at line 7

最佳答案

您需要将子查询与Announcements表关联：

select *
from Announcements a
where exists (
        select 1
        from Categories_announcements ca
        inner join Categories_user cu on cu.category_id = ca.category_id
        where cu.user_id = 1
            and ca.announcement_id = a.id
        )

Demo