PhP搜索目錄通過文件擴展名

Question

我有一個腳本和一個函數在其中獲取該文件夾中的每個文件。如何編輯此代碼以使其僅獲取.png文件？

public function GetPreviews($category) 
    { 
     $directories = glob("assets/preview/$category/*"); 
     $directory = array(); 
     foreach($directories as $directories) 
     { 
      $directory[] = str_replace("assets/preview/$category/", "", $directories); 
     } 
     return $directory;  
    } 

Answer 1

由於您使用​​3210 - 剛剛過去的*

之後添加.png像這樣：

public function GetPngs($category) 
    { 
     $directories = glob("assets/preview/$category/*.png"); 
     $directory = array(); 
     foreach($directories as $directories) 
     { 
      $directory[] = str_replace("assets/preview/$category/", "", $directories); 
     } 
     return $directory;  
    } 

Answer 2

你缺少你的代碼 'png格式' 只是改變

$directories = glob("assets/preview/$category/*"); 

至

$directories = glob("assets/preview/$category/*.png"); 

Answer 3

將.png添加到glob（「assets/preview/$ category/*。png」）;

public function GetPreviews($category) 
    { 
     $directories = glob("assets/preview/$category/*.png"); 
     $directory = array(); 
     foreach($directories as $directories) 
     { 
      $directory[] = str_replace("assets/preview/$category/", "", $directories); 
     } 
     return $directory;  
    }