【问题标题】:MySQL / PDO - Select from other table (Join statement)MySQL / PDO - 从其他表中选择(加入语句)
【发布时间】:2015-02-25 15:39:21
【问题描述】:

我有一个 PDO 语句,它从一些不同的表中选择一些数据。虽然,我不确定如何从另一个表中选择数据:

SELECT 
        c.forum_id as category_id,
        c.forum_name as category_name,
        t.forum_id as id,
        t.forum_name as name,
        t.forum_desc as description,
        (SELECT COUNT(*) FROM forum_topics WHERE forum_id=t.forum_id AND topic_deleted=0) as topics_count,
        (SELECT COUNT(*) FROM forum_posts WHERE forum_id=t.forum_id AND post_deleted=0) as posts_count,
        (SELECT COUNT(*) FROM forum_posts WHERE topic_id=lp.topic_id AND post_deleted=0) as last_post_count,
        lp.topic_id as last_post_topic_id,
        lp.topic_title as last_post_topic_title,
        lp.post_time as last_post_time,
        lp.username as last_post_username
    FROM forum_cats as t
    JOIN forum_cats as c on c.forum_id = t.forum_type_id
    left join (
        SELECT 
            ft.topic_id,
            ft.title as topic_title,
            tmp.post_time,
            u.username,
            fp.forum_id
        FROM
            forum_posts fp
            join forum_topics ft on ft.topic_id = fp.topic_id
            join users u on u.id = fp.userid
            join (
                select forum_id, max(`post_time`) `post_time`
                from forum_posts fp
                where fp.post_deleted = 0
                group by forum_id
                ) as tmp on (fp.forum_id = tmp.forum_id and fp.post_time = tmp.post_time)
        where post_deleted = 0 and ft.topic_deleted = 0
    ) as lp on lp.forum_id = t.forum_id
    where t.forum_active = 1 and c.forum_active = 1
    order by category_id, t.forum_id
");

现在,我想选择avatar.users 列,其中username.users = last_post_usernameavatar 列,来自users 表)。

我完全不知道从哪里开始。我应该这样写吗?

(SELECT avatar FROM users WHERE username=last_post_username)

但那行不通。

感谢任何帮助。

【问题讨论】:

    标签: php mysql pdo left-join inner-join


    【解决方案1】:

    您已经拥有基于“u.id = fp.userid”的连接的“last_post_username”。您只需要检索“lp.avatar”

    这是我的答案:

    SELECT 
            c.forum_id as category_id,
            c.forum_name as category_name,
            t.forum_id as id,
            t.forum_name as name,
            t.forum_desc as description,
            (SELECT COUNT(*) FROM forum_topics WHERE forum_id=t.forum_id AND topic_deleted=0) as topics_count,
            (SELECT COUNT(*) FROM forum_posts WHERE forum_id=t.forum_id AND post_deleted=0) as posts_count,
            (SELECT COUNT(*) FROM forum_posts WHERE topic_id=lp.topic_id AND post_deleted=0) as last_post_count,
            lp.topic_id as last_post_topic_id,
            lp.topic_title as last_post_topic_title,
            lp.post_time as last_post_time,
            lp.username as last_post_username,
            lp.avatar
        FROM forum_cats as t
        JOIN forum_cats as c on c.forum_id = t.forum_type_id
        left join (
            SELECT 
                ft.topic_id,
                ft.title as topic_title,
                tmp.post_time,
                u.username,
                u.avatar,
                fp.forum_id
            FROM
                forum_posts fp
                join forum_topics ft on ft.topic_id = fp.topic_id
                join users u on u.id = fp.userid
                join (
                    select forum_id, max(`post_time`) `post_time`
                    from forum_posts fp
                    where fp.post_deleted = 0
                    group by forum_id
                    ) as tmp on (fp.forum_id = tmp.forum_id and fp.post_time = tmp.post_time)
            where post_deleted = 0 and ft.topic_deleted = 0
        ) as lp on lp.forum_id = t.forum_id
        where t.forum_active = 1 and c.forum_active = 1
        order by category_id, t.forum_id
    ");
    

    【讨论】:

    • 我如何才能在 php 中使用“头像”? $row['头像']?
    猜你喜欢
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    相关资源
    最近更新 更多