我有具有hasMany关系的Posts and Comments模型:
public function comments()
{
return $this->hasMany(Posts::class, 'posts_id', 'id');
}
在我的控制器中,我需要获取所有已发布的帖子(is_published = 1),其中包含所有已发布的评论,并且至少具有1个已发布的评论:
$dbRecords = Posts::all()->whereStrict('is_published', 1);
$posts = [];
foreach ($dbRecords as $post) {
if (count($post->comments()) === 0) {
continue;
}
foreach ($post->comments() as $comment) {
if ($comment->is_published === 1) {
$posts[] = $post;
continue(2); // to the next post
}
}
}
但是,这样的解决方案很难看。另外,我将获得所有已发布的帖子,机智已发布和未发布的评论,因此我将不得不再次在Resource中过滤评论。
我发现的另一种解决方案-使用原始查询:
$dbRecords = DB::select("SELECT posts.*
FROM posts
JOIN comments ON posts_id = posts.id
WHERE posts.is_published = 1
AND comments.is_published = 1
HAVING count(posts.id) > 0;");
$users = array_map(function($row) { return (new Posts)->forceFill($row); }, $dbRecords);
但是它不能解决需要过滤Resource中未发表评论的问题。
答案 0 :(得分:1)
Eager loading
来解决在laravel中使用n+1
的{{1}}查询问题。with
或has
函数来querying relationship existence。如果您是这样的话。
whereHas
答案 1 :(得分:0)
如何雄辩
$posts = Post::query()->where('is_published', 1)->with(['comments' => function ($query) {
$query->where('is_published', 1);
}])->get();