Question

UserProductsController

class UserProductsController extends Controller
{
    public function index()
    {
        $products = Product::get();
        return view ('products')->with(compact('products'));
      
    }
      
    public function product_categories()
    {
        $categories = Category::all();
        return view ('categories')->with(compact('categories'));
      
    }
  
}

products_table

 public function up()
    {
        Schema::create('products', function (Blueprint $table) {
            $table->bigIncrements('id');
            $table->string('prod_name');
            $table->string('prod_brand')->nullable();
            $table->unsignedBigInteger('cat_id');
            $table->string('prod_image_path')->nullable();
            $table->timestamps();

            $table->foreign('cat_id')
            ->references('id')
            ->on('categories')
            ->onDelete('cascade');
        });
    }

categories_table

   public function up()
        {
            Schema::create('categories', function (Blueprint $table) {
                $table->bigIncrements('id');
                $table->string('cat_name');
                $table->string('cat_image_path')->nullable();
                $table->string('cat_description')->nullable();
                $table->timestamps();
            });
        }

产品型号

class Product extends Model
{
  

    public function category()
    {
        return $this->belongsTo('App\Category','category_id');
    }
 
    
}

类别模型

class Category extends Model
{
   public function category()
   {
       return $this->hasMany('App\Product');
   }
}

我的路线是

Route::get('/All_Products', 'UserProductsController@index')->name('All_Products');
Route::get('/product_categories', 'UserProductsController@product_categories')->name('categories');

如何获得相同类别的所有产品？由于这是我的第一个项目，因此我将花费更多的时间。但是对我没有任何作用。有人可以引导我吗？

Answer 1

假设您正确设置了关系（不是这样）

您有几种使用口才的方法：

$products = Category::findOrFail($categoryId)->products;

$products = Product::where('category_id', $categoryId)->get();

$products = Product::whereHas('category', function ($query) use ($categoryId) {
    $q->where('id', $categoryId);
})->get();

列举一些。

Laravel 7.x Docs - Eloquent - Retrieving Single Models / Aggregates findOrFail

Laravel 7.x Docs - Eloquent - Relationships - Relationship Methods vs Dynamic Properties

Laravel 7.x Docs - Eloquent - Relationships - Querying the Existence of Relationships whereHas

Answer 2

首先，您没有正确定义您的关系。应该是这样的：

class Product extends Model
{
    public function category()
    {
        return $this->belongsTo('App\Category');
    }
}

class Category extends Model
{
   public function products()
   {
       return $this->hasMany('App\Product');
   }
}

然后在产品迁移文件中，cat_id应该重命名为category_id。这样，您无需在关系上指定外键。

我假设您要列出属于特定类别的所有产品。您可以使用路由模型绑定轻松地做到这一点。在这种情况下，您的路线应类似于：

Route::get('categories/{category:id}/products', [CategoryController::class, 'products']);

然后在您的控制器中：

use App\Category;

class CategoryController extends Controller
{
    public function products(Category $category)
    {
        $category->load('products');

        return view('products')->withCategory($category);
    } 
}

您可以通过以下方式在刀片视图中访问产品列表：$category->products

Answer 3

您需要对Category模型进行调整，因为Category有很多Products。就目前而言，关系的命名并不能反映出这一点

class Category extends Model
{
   public function products()
   {
       return $this->hasMany('App\Product');
   }
}

然后您可以像这样通过Category模型访问产品。

$categories = Category::with('products')->all();

$categories->each(function($category) {
    $products = $category->products;
    
    // Dump & Die a collection of products
    dd($products);
});

注意：我渴望使用with()方法加载该关系，这只是为了防止n+1查询。有关急切和延迟加载的更多信息，请参见文档。

Answer 4

您可以做类似的事情，

$products = product::with('categories')->get();
foreach($products as $product)
{
    foreach($product->categories as $category)
    {
        echo $category->name;
    }
}
$categories = Category::with('products')->get();

$category = Category::with('products')->find($category_id);

Laravel Eloquent如何根据类别ID获取所有产品

4 个答案: