在我的Laravel应用程序中,我有一个类别列表页面。当用户单击某个子类别时,我希望列出所有产品,并对该结果使用分页。现在,我已经列出了与该子类别相关的所有产品,使用子类别ID:
public function subcategoryListing($slug){
$products = Subcategory::find($idofSubcat)->products;
return view('pages.subcategorylisting')
->with(array(
'products' => $products,
));
}
这个结构包含三个类:Category、Subcategory和Products。它们的声明如下:
<?php
namespace App;
use Illuminate'Database'Eloquent'Model;
use Illuminate'Database'Eloquent'Relations'Relation;
use App'Subcategory;
class Category extends Model
{
protected $table = 'category';
public $timestamps = false;
public function subCategory(){
return $this->hasMany('App'Subcategory', 'category_id');
}
}
子类
<?php
namespace App;
use Illuminate'Database'Eloquent'Model;
class Subcategory extends Model
{
protected $table = 'subcategory';
public $timestamps = false;
public function products(){
return $this->hasMany('App'Products');
}
}
产品<?php
namespace App;
use Illuminate'Database'Eloquent'Model;
class Products extends Model
{
protected $table = 'products';
}
对于每个模型类,我有一个表,结构如下:
Category
- id
- category_name
SubCategory
- id
- category_id
- subcategory_name
Products
- id
- subcategory_id
- product_title
- description
- price
我想要在我的页面中对从查询中检索到的结果进行分页。是否有更好的方法来获取与subcategory
相关联的products
并对它们进行分页?
在Eloquent (Laravel的ORM)中,当你调用关系作为属性($subCategory->products
)时,它会根据关系类型(属于,有许多,…)返回相关对象或对象集合。相反,如果您将它作为一个函数($subCategory->products()
)调用,您将得到一个QueryBuilder
实例。
请参阅http://laravel.com/docs/5.1/eloquent-relationships#querying-relations, 关系方法与动态属性一节了解更多详细信息。
无论如何,使用关系方法,您可以为您的集合调用paginate()
。然后,考虑到这一点,您可以稍微更改代码以得到您想要的:
public function subcategoryListing($slug) {
// I'm supposing here that in somewhere before
// run the query, you set the value to $idofSubcat
// variable
$products = Subcategory::find($idofSubcat)->products()->paginate();
return view('pages.subcategorylisting')
->with(array(
'products' => $products,
));
}
我认为,创建两个类别表不太正确,最好使用下一个:
表类别id category_name parent_id(nullable)
产品和 id category_id product_title description price
它更有用,你可以删除一个子类别模型并在类别中执行所有操作。
namespace App;
use Illuminate'Database'Eloquent'Model;
class Category extends Model
{
protected $table = 'category';
public $timestamps = false;
public function subCategory(){
return $this->belongsToMany('App'Category', 'categories', 'id', 'parent_id');
}
public function products(){
return $this->hasMany('App'Products');
}
}
和产品型号
<?php
namespace App;
use Illuminate'Database'Eloquent'Model;
class Products extends Model
{
protected $table = 'products';
public function category() {
return $this->belongsTo('App'Category')
}
}
,则可以得到查询结果
public function subcategoryListing($slug){
$products = Category::find($idofSubcat)->products;
return view('pages.subcategorylisting')
->withProducts($products); // it's a magic))
}
但是,存在着一个不漂亮的东西。你真的确定产品只属于一个类别吗?
我真的不明白你最初的一些功能:
// $slug is not being used anywhere within the function
public function subcategoryListing($slug){
// $idOfSubcat isn't passed to this function so will throw an error
$products = Subcategory::find($idofSubcat)->products;
return view('pages.subcategorylisting')
->with(array(
'products' => $products,
));
}
如果您想要做的是对属于特定subcategory_id
的Products
进行分页,如果您有subcategory_id
,则以下代码将起作用:
$products = Product::where('subcategory_id', $idofSubcat)->paginate();
然后你可以把这个分页集合返回给你的视图,就像你已经做的那样:
return view('pages.subcategorylisting')
->with(compact('products'));