排序文件以创建所有文件的对象

Question

我想从目录下的文件夹中获取所有图像，并将它们转换为对象。文件夹名称是产品ID，文件名只是数字。

我有这样的代码

  $dir = "./public/img/product/"; 
      $scandir = scandir($dir); 
      $opendir = opendir($dir); 
      $allfiles = new stdClass(); 

      while (false !== ($dirname = readdir($opendir))) 
      { 
       if($dirname != "." && $dirname != "..") 
       { 
         $allfiles->$dirname = $dirname; 

       } 

      } 


      foreach($allfiles as $folder) 
      { 
        $dir = "./public/img/product/".$folder; 
        $opendir = opendir($dir);   
        while (false !== ($filename = readdir($opendir))) 
       { 
        if($filename != "." && $filename != "..") 
        { 
         echo 'folder: '.$folder . ', filname: '.$filename.' '; 
        } 
       } 
//  

      } 

，使

folder: 767, filname: 1.jpg folder: 767, filname: 2.jpg folder: 767, filname: 3.jpg folder: 768, filname: 1.jpg folder: 768, filname: 2.jpg folder: 769, filname: 1.jpg folder: 769, filname: 2.jpg folder: 769, filname: 3.jpg folder: 769.. 

我想是这样的，我想..

{files:{767:[ 1.jpg, 2.jpg ],768:[ 1.jpg, 2.jpg ]}} 

谢谢。

Answer 1

这里是解决方案

 $dir = "./public/img/product/"; 
     $scandir = scandir($dir); 
     $opendir = opendir($dir); 
     $allfiles = new stdClass(); 
     $files = array(); 

     while (false !== ($dirname = readdir($opendir))) 
     { 
      if($dirname != "." && $dirname != "..") 
      { 
        $allfiles->$dirname = $dirname; 

      } 

     } 


     foreach($allfiles as $folder) 
     { 
       $dir = "./public/img/product/".$folder; 
       $opendir = opendir($dir);   
       while (false !== ($filename = readdir($opendir))) 
      { 
       if($filename != "." && $filename != "..") 
       { 
        $files[$folder][] = $filename; 
       } 
      } 

     } 

     ksort($files); 
     foreach($files as $index => $folder){ 
      $files[$index] = sort($folder); 
     } 

     echo json_encode($files);