Question

我正在努力使用Laravel Query Builder构建以下嵌套复杂查询。我正在使用laravel 4.0。嵌套连接的主要问题。

SELECT clubs.id, clubs.slug, clubs.name as `club_name`, clubs.state, clubs.category, clubs.live, `m`.`total_members`, `m`.`online_members`, `m`.`offline_members`, `m`.`last_7days_members` as `total_members_last_7days`

            FROM clubs

            LEFT JOIN 
                (
                    SELECT id, club_id, count(*) as `total_members`, SUM(online_member = 1) AS online_member 

                    FROM members 

                    LEFT JOIN

                        (
                            SELECT posts.id, posts.member_id, posts.title, posts.body 
                            FROM posts 
                            WHERE posts.transaction_id IS NOT NULL

                        ) p on p.member_id = members.id
                    WHERE (`post_verfied` is null or `post_verfied` = 1) 
                        AND `members`.`deleted_at` is null
                    group by club_id
                ) m on m.club_id = clubs.id

            LEFT JOIN comments on comments.member_id = m.id

            WHERE clubs.`deleted_at` is null
            group by `clubs`.name

请帮助..

Answer 1

您可以使用DB:raw进行操作，如下所示：

DB:table('table1')->leftJoin(DB::raw('(select * 
                             from table2 
                             join table3 on table3.id = table2.id) as mytable'),
function($join)
{
    $join->on('table1.id', '=', 'mytable.id');
});

我知道这不是你想要的，但它是一个部分解决方案。调查我发现Laravel 4.x不支持。一年前在Laravel 3.x中有一个pull，但最终没有实现。无论如何，您都可以在Builder.php中查看使用查询构建器的所有选项。

当您无法使用Query Builder或Eloquent ORM执行复杂查询时，也可以使用DB:statement。

嵌套加入Laravel 4.0

1 个答案: