Question

如何使用变量作为新属性为对象提供新属性？

以下为我提供了所需的属性对象：

switch ($property['property_type']):
    case 'Residential':
        $property = $this->property
                         ->join('residential', 'property.id', '=','residential.property_id')
                         ->join('vetting', 'property.id', '=', 'vetting.property_id')
                         ->where('property.id', $id)
                         ->first();

        $property['id'] = $id;
        break;
    default:
        return Redirect::route('property.index');
        break;
endswitch;

以下为我提供了属性和值的列表：

$numeric_features = App::make('AttributesController')->getAttributesByType(2);

以下是问题，如何将$numeric_features中的每一个动态添加到属性对象？

foreach ($numeric_features as $numeric_feature) {
    ***$this->property->{{$numeric_feature->name}}***=$numeric_feature->value;
}

Answer 1

查看http://php.net/manual/en/function.get-object-vars.php

$property_names = array_keys(get_object_vars($numeric_features));

foreach ($property_names as $property_name) {
   $property->{$property_name} = $numeric_features->{$property_name};
}

并检查此eval结果，它将一个对象的属性添加到另一个对象： https://eval.in/517743

$numeric_features = new StdClass;
$numeric_features->a = 11;
$numeric_features->b = 12;

$property = new StdClass;
$property->c = 13;

$property_names = array_keys(get_object_vars($numeric_features));

foreach ($property_names as $property_name) {
   $property->{$property_name} = $numeric_features->{$property_name};
}
var_dump($property);

结果：

object(stdClass)#2 (3) {
  ["c"]=>
  int(13)
  ["a"]=>
  int(11)
  ["b"]=>
  int(12)
}

从另一个分配一个对象属性

1 个答案: