Laravel 8 路由到控制器。 SEO友好的URL结构

Question

我试图弄清楚如何在 Laravel 8 项目中实现特定的 URL 结构以及实现这一目标的必要途径。我想要的是：

// Example urls to listings in the business directory.
// These urls should be routed to the directory controller.
www.domain-name.com/example-business-name-d1.html
www.domain-name.com/example-business-name-d15.html
www.domain-name.com/example-business-name-d100.html
www.domain-name.com/example-business-name-d123.html
www.domain-name.com/example-business-name-d432.html

// Example urls to articles/posts in the blog.
// These urls should be routed to the posts controller.
www.domain-name.com/example-post-name-p5.html
www.domain-name.com/example-post-name-p11.html
www.domain-name.com/example-post-name-p120.html
www.domain-name.com/example-post-name-p290.html
www.domain-name.com/example-post-name-p747.html

// We want to avoid the more traditional:
www.domain-name.com/directory/example-business-name-1.html
www.domain-name.com/blog/example-post-name-5.html

这是因为我们不希望每个列表或博客文章的 URL 中都包含字符串“目录”或“博客”。没有它，搜索引擎结果会更好。

到目前为止，我在 web.php 路由文件的底部使用了一个包罗万象的路由 {any} 来“捕捉所有”的路由。然后我操纵路径提供的字符串以从 url 的末尾获取 ID 和单个字符标记。然后我有了这 2 个变量，但可以弄清楚如何将它们传递给正确的控制器！

还是我真的很笨，有更好的方法来实现这一点？

Route::get('{any}', function($any = null){

    // Break up the url into seperate parts.
    $pieces = explode("-", $any);
    $pieces = array_reverse($pieces);
    $piece =  $pieces[0];

    // Remove the .html
    $piece = substr($piece, 0, -5);

    // Get the two parts of the identifier.
    $id = substr($piece, 1);
    $token = substr($piece, 0, 1);

    // Call correct controller based on the token.
    switch ($token) {
        case "d":
            // HERE I WANT TO PASS THE ID ON TO THE SHOW ACTION OF THE DIRECTORY CONTROLLER 
        break;
        case "p":
            // HERE I WANT TO PASS THE ID ON TO THE SHOW ACTION OF THE POSTS CONTROLLER 
        break;
        default:
            return abort(404);
        break;
    }

});

Answer 1

您可以使用正则表达式匹配 slug

Route::get('/{any}', 'YourController@methodName')->where(['any' => '.*(-d(.*?)\.).*']);

与p重复

然后，当您在控制器方法中获取 $site 时，您可以使用正则表达式来获取站点。

public function methodName($site)
{
    preg_match('/.*(-(d(.*?))\.).*/', $site, $parts); //or something similar, $parts[2] will have what you want
}

或

这会给你的控制器方法 d{number} 或 p{number}

Route::get('/{site}', function($site) {
    $code = preg_match('/.*(-(d(.*?)|p(.*?))\.).*/', $site, $parts) ? $parts[2] : null;

    $controllerName = 'ControllerA';
    if(isset($code) && !is_null($code) && Str::contains($code, 'p')) {
        $controllerName = 'ControllerB';
    }

    $controller = app()->make('App\Http\Controllers\Application\\' . $controllerName);

    return $controller->callAction('methodName', $params = ['code' => $code]);
})->where(['site' => '.*(-(d|p)(.*?)\.).*']);

Answer 2

我会将路径拆分为 2 个变量（$slug 和 $id）并直接将其传递给控制器​​。

Route::get('{slug}-d{id}.html', 'DirectoryController@show')
    ->where(['slug' => '([a-z\-]+)', 'id' => '(\d+)']);

Route::get('{slug}-p{id}.html', 'PostController@show')
    ->where(['slug' => '([a-z\-]+)', 'id' => '(\d+)']);

在你的控制器中

class DirectoryController
{
    public function show(string $slug, int $id) {}
}

class PostController
{
    public function show(string $slug, int $id) {}
}

Answer 3

我可以看到两种实现此结果的方法：

创建中间控制器

Route::get('{path}', 'CheckPathController@redirect')

然后在您的CheckPathController 中进行所有检查并调用正确的控制器操作：

public function redirect(Request $request, $path) {
  // Your checks on $path, extract $id and content type

  if($isPost) {
     $controller = resolve(PostController::class);
     return $controller->show($request, $id);
  }

  if($isBusiness) {
     $controller = resolve(BusinessController::class);
     return $controller->show($request, $id);
  }

  // No matches, error 404
  abort(404);
}

复杂的正则表达式

见：https://laravel.com/docs/8.x/routing#parameters-regular-expression-constraints

我不是正则表达式大师，这应该是匹配任何{word}-{word}-...-p{id}.html 模式的基本方法，但它会在出现意外字符时中断

Route::get('{path}', 'PostController::show')
   ->where(['path' => '([\w]*-)*p[0-9]+\.html$']);

Route::get('{path}', 'BusinessController::show')
   ->where(['path' => '([\w]*-)*d[0-9]+\.html$']);

请注意，在这种情况下，您的控制器将收到 pull $path 字符串，因此您需要在此处提取 id。