【发布时间】:2010-07-27 23:51:27
【问题描述】:
我不确定这个问题,但看看:
我有 3 张桌子:
- 类别:ID、图标、名称;
- 子类别:id、name、description、category_id;
- 主题:id、title、text、author、created、subcategory_id;
现在我正在尝试获取此信息(查询):
$String = "SELECT
categories.id AS catid,
categories.icon AS caticon,
categories.name AS catname,
subcategories.id AS scatid,
subcategories.name AS scatname,
subcategories.description AS scatdescription,
subcategories.category_id AS scatcatid,
COUNT(topics.id) AS tid,
topics.title AS ttitle,
topics.author AS tauthor,
topics.created AS tcreated
FROM
categories
LEFT JOIN
subcategories
ON
subcategories.category_id = 1
LEFT JOIN
topics
ON
subcategories.id = topics.subcategory_id
GROUP BY
categories.id";
结果:
5 中的类别显示 5 - OK, 4 个子类别中的第一个类别仅显示 1 个。
也许查询太长了?感谢您的任何回答。
【问题讨论】:
-
在第一次加入时,您已经写了:subcategories.category_id = 1 ...据我了解,这将仅加入类别 1 的所有子类别 ...
标签: mysql