【发布时间】:2015-08-03 22:32:34
【问题描述】:
花了8个小时检查这个问题,但没有任何结果,有人能说什么错了吗?
>SELECT id FROM posts AS post_id;SET categories.id AS category_id;
>SELECT title, contents, date_posted, categories, name FROM posts INNER JOIN
>categories ON categories, id = posts, cat_id ORDER BY id DESC
并得到一个错误 $ #1193 - 未知系统变量 'id'
我也尝试过这样的事情:
>SELECT id FROM posts AS post_id;
>SELECT id FROM categories AS category_id;
>SELECT COALESCE (title, contents, date_posted, categories, name)
>FROM posts INNER JOIN categories ON categories, id = posts, cat_id
>ORDER BY id DESC
还有很多其他无法正常工作的员工。
工作查询:
>SELECT posts . p_id AS post_id, categories . c_id AS category_id,
>title, contents, date_posted, categories . name
>FROM posts
>INNER JOIN categories ON categories . c_id = posts . cat_id
>ORDER BY posts.p_id DESC;
【问题讨论】:
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抱歉英语不好)
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ON categories, id??你的意思是ON categories.id -
不,我的意思是必须加入 2 个表 ID,来自“类别”和“帖子”,以及它的 ID。
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对不起,你说得对)谢谢)
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你能把
SHOW CREATE TABLE的结果粘贴到每个表吗?
标签: mysql