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例如,我有一个包含 100 个“人”节点的 XML 文件,我想要前 30 个。或者可能是 51 - 100。有没有办法使用 e4x 语法来返回一个 XMLList?
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());