从多个表中选择列？

Question

我需要从3个单独的表中检索数据，因为我已经向正在使用的数据库中添加了另一个表。

我的布局目前看起来像这样：

 _____________       _____________      ___________      ___________
| users       |     |    links    |    |users_links|    | tags      |
|-------------|     |-------------|    |-----------|    |-----------|
| id          |     | id          |    | id        |    | id        |
| name        |     | long_url    |    | link_id   |    | user_id   |
| username    |     | short_url   |    | user_id   |    | link_id   |
| email       |     | user_id     |    | privacy   |    | tag       |
| password    |     | page_title  |    | notes     |    |           |
| created_at  |     | updated_at  |    |           |    |           |
| updated_at  |     | clicks      |    |-----------|    |-----------|
---------------     | page_title  |
                    --------------            

以前，我是从两个表中收集数据的（效果很好），但是在感觉到需要允许标记之后，这需要一个额外的表。

我的原始查询如下所示：

    public static function previouslyShortened($username)
    {
            //Gets all related info for the info for member page
            $query = DB::select("select users.id as user_id,
                    links.id,
                    links.long_url,
                    links.short_url,
                    links.updated_at,
                    links.clicks,
                    links.page_title,
                    users_links.privacy,
                    users_links.notes,
                    from users
                    right outer join users_links on users.id = users_links.user_id
                    join links on links.id = users_links.link_id
                    where users.username = '$username'
                    ". (Auth::check() && Auth::user()->username == $username ? "" : "and users_links.privacy = 0")
            );

            $result = array('username' => $username, 'links' => $query);
            return $result;
    }

我还希望在查询中包含表tags中的tags.id和tags.tag ，并将其作为正在创建的对象的一部分返回。

我尝试通过尝试创建另一个正确的外部联接来尝试此操作，但出现错误：

Syntax error or access violation - Not unique table alias. 

我怎样才能做到这一点？

---

编辑：

请求我尝试修复，因此这里是：

    public static function previouslyShortened($username)
    {
            //Gets all related info for the info for member page
            $query = DB::select("select users.id as user_id,
                    links.id,
                    links.long_url,
                    links.short_url,
                    links.updated_at,
                    links.clicks,
                    links.page_title,
                    users_links.privacy,
                    users_links.notes,
                    tags.id,
                    tags.tag
                    from users
                    right outer join users_links on users.id = users_links.user_id
                    join links on links.id = users_links.link_id
                    right outer join tags.user_id on users_links.id = tags.user_id
                    join tags on users_links.user_id = tags.user_id
                    where users.username = '$username'
                    ". (Auth::check() && Auth::user()->username == $username ? "" : "and users_links.privacy = 0")
            );

            $result = array('username' => $username, 'links' => $query);
            return $result;
    }

Answer 1

要消除错误，您可以尝试以下代码

public static function previouslyShortened($username)
{
        //Gets all related info for the info for member page
        $query = DB::select("select users.id as user_id,
                links.id,
                links.long_url,
                links.short_url,
                links.updated_at,
                links.clicks,
                links.page_title,
                users_links.privacy,
                users_links.notes,
                tags.id,
                tags.tag
                from users
                left join users_links on users.id = users_links.user_id
                left join links on links.id = users_links.link_id
                left join tags on users.id = tags.user_id
                where users.username = '$username'
                ". (Auth::check() && Auth::user()->username == $username ? "" : "and users_links.privacy = 0") . " group by links.id"
        );
        $result = array('username' => $username, 'links' => $query);
        return $result;
}

Answer 2

在我们的聊天中，这是from语句：

from users 
inner users_links on users.id = users_links.user_id 
inner join links on links.id = users_links.link_id 
left join tags as t1 on users_links.id = t1.user_id 
left tags as t2 on users_links.user_id = t2.user_id