How to write this (left join, subquery ) in Laravel 5.1?(如何在 Laravel 5.1 中编写这个(左连接、子查询)?)
问题描述
如何在 Laravel 5.1 中编写此查询:
How to write this query in Laravel 5.1:
SELECT p.id, p.title, p.created_at, p.updated_at, u.name, COALESCE(c.comments_count, 0) AS comments_count, COALESCE(pl.status_sum, 0) AS status_sum
FROM posts p
LEFT OUTER JOIN users u ON u.id = p.user_id
LEFT OUTER JOIN (
SELECT pl.post_id, SUM(pl.status) AS status_sum
FROM postslikes pl
GROUP BY pl.post_id
) pl ON pl.post_id = p.id
LEFT OUTER JOIN (
SELECT c.post_id, COUNT(*) as comments_count
FROM comments c
GROUP BY c.post_id
) c ON c.post_id = p.id ORDER BY comments_count DESC
我需要它来进行分页.我可以毫无问题地执行这个查询,但是手动分页器总是给出相同的结果:http://laravel.com/docs/5.1/pagination#manually-creating-a-分页器与此处相同的问题:http://laravel.io/forum/07-22-2015-laravel-51-manual-pagination-not-working-as-expected
I need it for Pagination. I can perform this query raw without any problems but the manually paginator gives always the same results: http://laravel.com/docs/5.1/pagination#manually-creating-a-paginator The same problem as here: http://laravel.io/forum/07-22-2015-laravel-51-manual-pagination-not-working-as-expected
我的尝试没有成功:
DB::table( 'posts' )
->select( 'posts.id', 'posts.title', 'posts.created_at', 'posts.updated_at', 'users.name', DB::raw( 'COALESCE( comments.body, 0 ), COALESCE( postslikes.status, 0 )' ) )
->leftJoin( 'users', 'users.id', '=', 'posts.user_id' )
->leftJoin( DB::raw( 'SELECT postslikes.post_id, SUM( postslikes.status ) FROM postslikes GROUP BY postslikes.post_id' ), function( $join )
{
$join->on( 'postslikes.post_id', '=', 'post.id' );
})
->leftJoin( DB::raw( 'SELECT comments.post_id, COUNT(*) FROM comments GROUP BY comments.post_id' ), function( $join )
{
$join->on( 'comments.post_id', '=', 'post.id' );
})
->get();
我认为问题在于comments_count 和status_sum?谢谢!
I think the problem is comments_count and status_sum? Thanks!
推荐答案
要在 Laravel 的查询构建器中使用子查询,您应该将其添加到连接中,如下所示:
To use subqueries with Laravel's query builder, you should add it to the join as follows:
->leftJoin(DB::raw("(SELECT [...]) AS p"), 'p.post_id', '=', 'posts.id')
最好为计算字段创建别名,就像您在原始查询中所做的那样:
It's also better to create an alias for calculated fields, as you did in your raw query:
COUNT(*) AS count
尽管发生了这些变化,除非我错了,否则您可以从简化查询开始.以这种方式删除子查询:
Despite this changes, unless I'm wrong, you can start by making your query simpler. Drop the subqueries, this way:
SELECT
p.id,
p.title,
p.created_at,
p.updated_at,
u.name,
COUNT(c.id) AS comments_count,
COALESCE(SUM(pl.status), 0) AS status_sum
FROM
posts p
LEFT OUTER JOIN
users u
ON
u.id = p.user_id
LEFT OUTER JOIN
postslikes pl
ON
pl.post_id = p.id
LEFT OUTER JOIN
comments c
ON
c.post_id = p.id
ORDER BY
comments_count DESC
GROUP BY
p.id
然后,通过这个新查询,您可以使用 Laravel 来构建它:
Then, with this new query, you can use Laravel to build it:
DB::table('posts')
->select([
'posts.id',
'posts.title',
'posts.created_at',
'posts.updated_at',
'users.name',
DB::raw('COUNT(comments.id) AS comments_count'),
DB::raw('COALESCE(SUM(postslikes.status), 0) AS status_sum'),
])
->leftJoin('users', 'users.id', '=', 'posts.user_id')
->leftJoin('comments', 'comments.post_id', '=', 'posts.id')
->leftJoin('postslikes', 'postslikes.post_id', '=', 'posts.id')
->orderBy('comments_count', 'DESC')
->groupBy('posts.id')
->get();
请注意,我假设您的 comments
表中有一个名为 id
的列作为主键.
Note that I'm assuming you have a column named id
in your comments
table that is the primary key.
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本文标题为:如何在 Laravel 5.1 中编写这个(左连接、子查询)?
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