Groovy提取另一个文件中的XML节点列表

Question

我有xml文件，其中包含数千个xml节点。在我的groovy代码中，我想提取其中的N个并推送到另一个文件中。我如何用groovy做到这一点？ 一些样本列表：

def ibr = new File('/Users/alex/Downloads/temp.xml');
def ibrParser = new XmlParser().parseText(ibr.getText());


def groups = [];
int current = 0;

ibrParser.each { group ->
    if (current <= 100) {
        groups<<group
    }
    current++
}
//how to store groups as xml into another file?

Answer 1

以下代码帮助了我：

int current = 0;

def nodeToSerialize;

ibrParser.each { renewalQuote ->
    if (current <= 100) {
        if (nodeToSerialize == null) {
            nodeToSerialize = renewalQuote;
        } else {
            nodeToSerialize.append(renewalQuote)
        }
    }
    current++
}


XmlUtil.serialize(nodeToSerialize, new FileOutputStream("/Users/alex/Downloads/100Cases.xml"))