Question

我有多条路线模型绑定路线，用于指导用户购物＆gt; 类别＆gt;的产品即可。路线链接正常。

但是如果用户向URL添加垃圾值，Laravel不会抛出404错误。如果用户向URL添加任何额外字符，如何强制404错误？

路线

Route::get('/{shop_url}/{category_url}/{product_url}/buy', 'Controller@buy')->name('buy')->where(['shop_url', 'category_url', 'product_url' => '[\w\d\-]+(.*)']);

Route::get('/{shop_url}/{category_url}', 'Controller@view')->name('view')->where(['shop_url','category_url' => '[\w\d\-]+(.*)']);

Route::get('/{shop_url}', 'Controller@shop')->name('shop')->where('shop_url', '[\w\d\-]+(.*)');

控制器

public function shop($shop_url)
{
  $shop = Shop::where('shop_url', $shop_url)->firstorfail();
  return view ('shop', compact('shop'));
}

public function view($shop_url, $category_url)
{
  $shop = Shop::where('shop_url', $shop_url)->firstorfail();
  $category= Category::firstorfail();

  return view ('shop', compact('shop', 'category'));
}

public function buy($shop_url, $category_url, $product_url)

{
  $shop = Shop::where('shop_url', $shop_url)->firstorfail();
  $category= Category::where('category_url', $category_url)->firstorfail();
  $product = Product::firstorfail();

  return view ('shop', compact('shop', 'category', 'product'));
}

此处domain.com/shop-ca/clothing有效，如果用户输入domain.com/shop-ca/clothes，它也会显示相同的页面。在这里，我希望它显示404.我该怎么做？

Answer 1

首先，你应该检查你的slug，在你的代码中我看到这些行：

$category= Category::firstorfail();// it will get the first route in the database, not the needed category

所以，首先，尝试这样做：

public function view($shop_url, $category_url)
{
   $shop = Shop::where('shop_url', $shop_url)->firstOrfail();
   $category= Category::where('category_url', $category_url)->firstOrfail();// I actually don't know the name of your slug field, but if the category is wrong, laravel will return 404 page.

   return view ('shop', compact('shop', 'category'));
}

Answer 2

您已在$product_url功能中检查buy，并手动检查404错误。

Laravel - 防止额外的字符slug参数

2 个答案: