Question

我的查询设计如下：

 $tags = auth()->user()->tags->toArray();
    return $this->posts($request)->whereHas('tags', function ($q) use ($tags) {
      $q->where(function ($q) use ($tags) {
        foreach ($tags as $tag) {
          $q->orWhere('tags.tag', 'like', $tag["tag"]);
        }
      })->select(DB::raw('count(distinct tags.id)'));
    })->paginate(perPage());

SQL可以是：

select * from `posts` 
where `posts`.`deleted_at` is null 
and `expire_at` >= '2017-03-26 21:23:42.000000' 
and (
select count(distinct tags.id) from `tags` 
inner join `post_tag` on `tags`.`id` = `post_tag`.`tag_id` 
where `post_tag`.`post_id` = `posts`.`id` 
and (`tags`.`tag` like 'PHP' or `tags`.`tag` like 'pop' or `tags`.`tag` like 'UI')
) >= 1

但是我需要按照帖子中的标签数量来排序结果，那么应该是这样的：

select p.*
from posts p
join (
select pt.post_id,
    count(distinct t.id) as tag_count
from tags t
inner join post_tag pt on t.id = pt.tag_id
where t.tag in ('PHP', 'pop', 'UI')
group by pt.post_id
) pt on p.id = pt.post_id
where p.deleted_at is null
and p.expire_at >= '2017-03-26 21:23:42.000000'
order by pt.tag_count desc;

是否可以在Laravel中创建它？如何完成查询？

Answer 1

使用withCount()方法：

->withCount(['tags' => function($q) {
    q->where.... // Put conditionals here if needed.
}])
->orderBy('tags_count', 'desc')

如果您想要计算关系中的结果数而不实际加载它们，您可以使用withCount方法，这会在生成的模型上放置{relation}_count列

Laravel - 查询生成器 - 按标签数量对结果进行排序

1 个答案: