根据特定模式生成最高n级的视图文件子文件

Question

我想使用存根文件中的（preg_match）从特定模式生成不超过n级的部分和子部分文件。让我们假设已生成带有布局文件的布局目录。我搜索了样式并制作了部分布局。如果现有文件中有部分样式，我想生成高达N级的子部分文件
到目前为止，我已经做到了

public function handle()
{
    $input = $this->getCommandInput();
    $path = $this->getDestenationPath($input->layoutDirectory);
    $destenationFile = $path . $input->layoutFileName;

    if ($this->alreadyExists($destenationFile)) {
        $this->error('The layout already exists!  To override the existing file, use --force option.');

        return false;
    }

    $stubName = $input->withoutValidation ? 'layout' : 'layout-with-validation';

    $stub= $this->getStubContent($stubName, $this->getTemplateName());

     $this->replaceApplicationName($stub, $input->appName)
        ->createFile($destenationFile, $stub)
        ->info('A Layout with Following Partials Have Been Crafted !');

         //generating partials//

         $partials=$this->replacePartialName($stub);


          foreach ($partials as $partial){

          print_r($partial);

        $partialContent=$this->getStubContent($partial);
        print_r($partialContent);
         $header=$this->replacePartialName($partialContent);
          print_r($header);

        $fileExtension=Helpers::postFixWith(trim($partial) ?: $partial, '.blade.php');
        $destenation=$this->getDestenationPath($input->layoutDirectory).$fileExtension;

        $this->replaceApplicationName($partialContent, $input->appName)
        ->createFile($destenation, $partialContent)
        ->info($partial.'.blade.php');

      }      

}