Question

我正在使用smugphp并完成了此代码块之上的所有登录内容。我想获取我在smugmug网站上创建的文件夹结构。不幸的是，这只给了我那个类别和文件夹树的叶子名称。似乎没有办法获得中间文件夹。

非常感谢任何帮助。

$categories=$f->categories_get();

$subcategories=$f->subcategories_getAll();

echo "CATEGORIES<br>";
for ($i=0;$i<count($categories);$i++){
  print_r($categories[$i]);
  echo "<br><br>";
}
echo "subcategories<br>";
for ($i=0;$i<count($subcategories);$i++){
  print_r($subcategories[$i]);
  echo "<br><br>";
}

Answer 1

我不知道这是最好还是最干净的解决方案，但是通过下面的代码，我可以欺骗smugmug给我想要的东西。我强制它进行深度优先搜索，然后忽略重复。我希望这有助于其他人，但如果有更好的解决方案，请告诉我。

<?php

require_once("phpSmug/phpSmug.php");

class Smugmug{
    private $f;
    private $folders;
    public function __construct(){
        $this->f=new phpSmug("APIKey=MfDv7qWG8VZt2tSK2DV7rZx10YXnu7UF","APIVer=1.2.2");
$this->f->login("APIKey=MfDv7qWG8VZt2tSK2DV7rZx10YXnu7UF","EmailAddress=EMAIL@gmail.com","Password=PASSWORD");
        $this->folders=[];
    }

    public function get_folders(){
        $this->folders=$this->_get_folders();
    }

    private function new_id($id,$folders){
        if (!isset($folders)){
            return False;
        }
        for ($i=0;$i<count($folders);$i++){
            if ($id==$folders[$i]['id']){
                return False;
            }
            if ($this->new_id($id,$folders[$i]['subfolders'])==False){
                return False;
            }
        }
        return True;
    }
    private function _get_folders($id=-1){
        $folders=[];
        if ($id==-1){
            $categories = $this->f->categories_get();
        } else {
            $categories = $this->f->subcategories_get("CategoryID=$id");
        }
        $k=0;
        for ($i=0;$i<count($categories);$i++){
            $folder['id']=$categories[$i]['id'];
            if ($this->new_id($folder['id'],$folders)){
                $folder['name']=$categories[$i]['Name'];
                $folders[$i]=$folder;
                $subfolders=$this->_get_folders($folder['id']);
                $folders[$i]['subfolders']=$subfolders;
            }
        }
        return $folders;
    }
    public function print_folders(){
        $this->_print_folders($this->folders);
    }
    private function _print_folders($folders,$prefix=""){
        for ($i=0;$i<count($folders);$i++){
            echo $prefix.$folders[$i]['name']."<br>";
            $this->_print_folders($folders[$i]['subfolders'],$prefix."-");
        }
    }
}

?>

如何通过smugmug api（phpsmug）获取文件夹结构

1 个答案: