laravel 5.1以多对多的关系获得每个类别的相关5个新闻

Question

我被困在这里从2-3小时开始尝试.

我有很多关系:

class Category extends Model
{
    public function news()
    {
        return $this->belongsToMany('App\News');
    }
}  

class News extends Model
{
    public function categories()
    {
        return $this->belongsToMany('App\Category');
    }
}

我正在尝试获取相关类别的最新5条新闻:

$front_categories = Category::with(array(
        'news'=>function($query){
        $query->where('publish','1')->orderBy('created_at', 'desc')->take(5);}))
       ->where('in_front', 1)->get();

上面的查询对我不起作用,它总共给出了五个结果,而不是每个类别的5个结果.

Answer 1

根据我对 Laravel 的了解，你可以尝试这样做。

\n\n

class Category {\n\n    public function recentNews()\n    {\n        return $this->news()->orderBy('created_by', 'DESC')\n                            ->take(5);\n    }\n}\n\n// Get your categories\n$front_categories = Category::where('in_front', 1)->get();\n\n// load the recent news for each category, this will be lazy loaded\n// inside any loop that it's used in.\nforeach ($front_categories as $category) {\n    $category->recentNews;\n}\n

\n\n

这与 L\xc3\xaa Tr\xe1\xba\xa7n Ti\xe1\xba\xbfn Trung 的答案具有相同的效果，并导致多个查询。它还取决于您是否重用此功能。如果它是一次性的，最好将其放在其他地方。其他方法也可以更加动态，例如创建一个返回类别集合的方法，您可以要求它提供某个数字：

\n\n

class CategoriesRepository {\n\n    public static function getFrontCategories(array $opts = []) {\n\n        $categories = Category::where('in_front', 1)->get();\n\n        if (!empty($opts) && isset($opts['withNewsCount'])) \n        {\n            foreach ($categories as $category) \n            {\n                $category->recentNews = static::getRecentNewsForCategory(\n                    $category->id,\n                    $opts['withNewsCount']\n                );\n            }\n        }\n\n        return $categories;\n    }\n}\n\n$front_categories = CategoriesRepository::getFrontCategories([\n    'withNewsCount' => 5\n]);\n

\n