在Groovy中对XML进行排序 [英] Sorting XML in Groovy

问题描述

我看过有关使用Groovy对XML进行排序的文档

I have looked at the documentation on sorting XML with Groovy

def records = new XmlParser().parseText(XmlExamples.CAR_RECORDS)
assert ['Royale', 'P50', 'HSV Maloo'] == records.car.sort{ it.'@year'.toInteger() }.'@name'

但是我想做的是对XML进行排序，然后返回已排序的xml字符串.我知道排序完成后就可以完全重建XML.

but what I am trying to do is sort the XML and then return the xml string sorted. I know I can completely rebuild the XML after i am done sorting.

我知道我可以对XML进行XML转换以对其进行排序

I know I can run an XML Transformation on the XML to get it sorted

def factory = TransformerFactory.newInstance()
def transformer = factory.newTransformer(new StreamSource(new StringReader(xslt)))
transformer.transform(new StreamSource(new StringReader(input)), new StreamResult(System.out))

但是我正在寻找一些Groovy魔术来使我更容易

BUT I was looking for some Groovy magic to make it easier for me

推荐答案

一种解决方案是直接替换 records 中的 car 列表.不确定是否存在更多魔法！

A solution is to replace directly the list of car within the records. Not sure if more magic exists!

records.value = records.car.sort{ it.'@year'.toInteger() }
println XmlUtil.serialize(records)

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