2009-10-14 11 views

回答

1
var list:XMLList = xml.person; 
var start:int = 10; 
var end:int = 40; 
var filteredList:XMLList = new XMLList(); 
for(i = start - 1; i < end; i++) 
    filteredList += new XML(XML(list[i]).toXMLString()); 
+1

VAR filteredList:XMLList中; 应该是 VAR filteredList:的XMLList =新的XMLList(); –

+0

糟糕,纠正。谢谢:) – Amarghosh

+1

错误:TypeError:错误#1086:appendChild方法只适用于包含一个项目的列表。替换for循环中的单个表达式有助于:var tempNode:XML = list [i]; filteredList + = tempNode; – eterps

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