Laravel 如何根据嵌套元素计数的结果筛选有说服力的关系结果
我当前正在以以下方式运行查询:Laravel 如何根据嵌套元素计数的结果筛选有说服力的关系结果,laravel,eloquent,Laravel,Eloquent,我当前正在以以下方式运行查询: $categories = Category::whereHas('amenities', function ($query) use ($selectedProperty) { $query->where('property_id', $selectedProperty); })->with(['amenities' => function($query) use($selectedProperty){
$categories = Category::whereHas('amenities', function ($query) use ($selectedProperty) {
$query->where('property_id', $selectedProperty);
})->with(['amenities' => function($query) use($selectedProperty){
$query->where('property_id',$selectedProperty)->orderByRaw('LENGTH(amenity_name)', 'ASC')->orderBy('amenity_name');
},'amenities.amenityValues','amenities.amenityValues.unitAmenityValues'])
->orderBy('category_name')
->get();
这里的关系是:
1 property = many amenities
1 amenity = many amenityValues
1 amenity Value = many unitAmenityValues
如果其unitAmenityValues
为空或在另一个短语中有0个count,我想过滤掉便利设施(或删除这些便利设施)。我只想要那些具有unitAmenityValues
的便利设施