php - 如何使用非法名称访问此对象属性？

Question

我正在使用某人编写的 PHP 类来与 BaseCamp API 交互。

我正在做的特定调用是检索待办事项列表中的项目，它工作正常。

我的问题是，我不确定如何仅访问返回的对象的 todo-items 属性。这是返回对象的 var_dump:

object(stdClass)[6]
  public 'completed-count' => string '0' (length=1)
  public 'description' => string 'Description String' (length=89)
  public 'id' => string '12345' (length=7)
  public 'milestone-id' => string '' (length=0)
  public 'name' => string 'Error Reports' (length=13)
  public 'position' => string '1' (length=1)
  public 'private' => string 'false' (length=5)
  public 'project-id' => string '58904' (length=7)
  public 'tracked' => string 'false' (length=5)
  public 'uncompleted-count' => string '1' (length=1)
  public 'todo-items' => 
    object(stdClass)[3]
      public 'todo-item' => 
        object(stdClass)[5]
          public 'completed' => string 'false' (length=5)
          public 'content' => string 'content string here' (length=133)
          public 'created-on' => string '2009-04-16T20:33:31Z' (length=20)
          public 'creator-id' => string '23423' (length=7)
          public 'id' => string '234' (length=8)
          public 'position' => string '1' (length=1)
          public 'responsible-party-id' => string '2844499' (length=7)
          public 'responsible-party-type' => string 'Person' (length=6)
          public 'todo-list-id' => string '234234' (length=7)
  public 'complete' => string 'false' (length=5)

如何访问该对象的 todo-items 部分？

Answer 1

<?php
$x = new StdClass();
$x->{'todo-list'} = 'fred';
var_dump($x);

所以，$object->{'todo-list'} 是子对象。如果你可以这样设置，那么你也可以这样读:

echo $x->{'todo-list'};

另一种可能性:

$todolist = 'todo-list';
echo $x->$todolist;

如果你想将它转换为一个数组，这可能更容易一点(即明显的 $ret['todo-list'] 访问)，这段代码几乎逐字从Zend_Config 会为你转换。

public function toArray()
{
    $array = array();
    foreach ($this->_data as $key => $value) {
        if ($value instanceof StdClass) {
            $array[$key] = $value->toArray();
        } else {
            $array[$key] = $value;
        }
    }
    return $array;
}