Laravel 嵌套路由

Question

我使用 laravel 7 并在数据库中存储以下树结构：

目录表：

category1
--category11
--category12
----category121
----category122
--category13

文章表：

news
--news1
--news2

基本的 Laravel 路由看起来像：

Route::get('category/{id}', 'categoryController@show');
Route::get('news/{id}', 'newsController@show');

但在这种情况下，每个目录的 URL 都需要“类别”url 的段，并且 每个新的 URL 都需要“news” url 的段

如何使用 Laravel Routes 路由以下 url：

http://sitename.com/category1
http://sitename.com/category1/category11
http://sitename.com/category1/category12/category121
http://sitename.com/news
http://sitename.com/news/news1

?

Answer 1

您需要一条“包罗万象”的路线（如果您希望 所有 路径可配置，请在所有其他路线之后注册）。

然后，您可以在斜杠上展开路径，检查每个元素是否是有效的类别 slug，并且也是前一个类别的子项。

// All other routes...

Route::get('/{category_path}', 'CategoryController@show')->where('category_path', '.*');

您也可以使用自定义路由绑定来做到这一点：

Route::bind('category_path', function ($path) {
    $slugs = explode('/', $path);

    // Look up all categories and key by slug for easy look-up
    $categories = Category::whereIn('slug', $slugs)->get()->keyBy('slug');

    $parent = null;   

    foreach ($slugs as $slug) {
        $category = $categories->get($slug);

        // Category with slug does not exist
        if (! $category) {
            throw (new ModelNotFoundException)->setModel(Category::class);
        }

        // Check this category is child of previous category
        if ($parent && $category->parent_id != $parent->getKey()) {
            // Throw 404 if this category is not child of previous one
            abort(404);
        }

        // Set $parent to this category for next iteration in loop
        $parent = $category;
    }

    // All categories exist and are in correct hierarchy
    // Return last category as route binding
    return $category;
});

然后您的类别控制器将接收路径中的最后一个类别：

class CategoryController extends Controller
{
    public function show(Category $category)
    {
        // Given a URI like /clothing/shoes,
        // $category would be the one with slug = shoes
    }
}