XMLのすべての子孫から特定の属性を削除するにはどうすればよいですか？

Question

私はID以外のすべての属性を削除し、XMLのすべてのノードを入力したい

var test:XML = new XML(<record id="5" name="AccountTransactions" 
    <field id="34" type="Nuber"/> 
    </record>); 

のようなXMLを持っています。このコードでは私はこれを行うことができません。あなたはループ以外のより良い解決策を提案できますか？

var atts:XMLListCollection = new XMLListCollection(test.descendants().attributes().((localName() != "id") && (localName() != "type"))); 
atts.removeAll(); trace(test) 

は、それはまだすべての属性を表示：/

Answer 1

var test:XML = new XML('<record id="5" name="AccountTransactions"><field id="34" type="Nuber" score="ded"/><field id="35" type="Nuber" score="sc"/></record>'); 
    var attributes:XMLList = test.field.@*; 
    var length:int = attributes.length(); 
    for (var i:int = 0; i < length; i++) { 
     (attributes[i].localName() != "id" && attributes[i].localName() != "type") ? [delete attributes[i], length--] : void; 
    } 
    trace(test); 

Answer 2

var xml:XML = new XML(
    <record id="5" name="AccountTransactions"> 
     <field id="34" type="Number"> 
      <test id="0"/> 
     </field> 
    </record>); 

//make array of attribute keys, excluding "id" and "type" 
var attributesArray:Array = new Array(); 
for each (var attribute:Object in xml.attributes()) 
{ 
    var attributeName:String = attribute.name(); 
    if (attributeName != "id" && attributeName != "type") 
    { 
     attributesArray.push(attributeName); 
    } 
} 

//loop through filtered attributes and remove them from the xml 
for each (var attributeKey:String in attributesArray) 
{ 
    delete xml.@[attributeKey]; 
    delete xml.descendants().@[attributeKey]; 
}