在gradle中使用特定id更新xml元素

我想在xml文件中做以下更改

My test.xml

<root>
    <display-name>Some Name</display-name>
    ....
    <resource-ref id='change'>
        <resource-name>Change this</resource-name>
    </resource-ref>
    <resource-ref id='modify'>
        <resource-name>Change this too</resource-name>
    </resource-ref>
</root>

某个名字
....
改变这个
把这个也换一下

我想在这个xml文件中做如下更改

<root>
    <display-name>Final Name</display-name>
    ....
    <resource-ref id='change'>
        <resource-name>After Change</resource-name>
    </resource-ref>
    <resource-ref id='modify'>
        <resource-name>After Modify</resource-name>
    </resource-ref>
</root>

task ("replace")<<{
def xmlSource = file(path/to/test.xml)
def xmlDest = file(path/to/destination)
def xmlParser = new XmlParser()
xmlParser.setFeature("http://apache.org/xml/features/disallow-doctype-decl", false)
def xmlRoot = xmlParser.parse(xmlSource)

xmlRoot.'display-name'[0].value = 'RTM16'
//Looking for something like this
//xmlRoot.'resource-ref'[@id='change'].'resource-name'[0].value = 'After Change'
//xmlRoot.'resource-ref'[@id='modify'].'resource-name'[0].value = 'After Modify'

def nodePrinter = new XmlNodePrinter(new PrintWriter(new FileWriter(xmlDest)))
nodePrinter.preserveWhitespace = true
nodePrinter.print(xmlRoot)
}

姓
....
变动后
修改后

问题的第一个答案几乎回答了我的问题。但我需要对具有特定id的元素进行特定更改，正如您所看到的

这看起来很简单。我试着寻找答案，但没有找到答案。感谢您的帮助

顺便说一句，我的gradle脚本是这样的

<root>
    <display-name>Final Name</display-name>
    ....
    <resource-ref id='change'>
        <resource-name>After Change</resource-name>
    </resource-ref>
    <resource-ref id='modify'>
        <resource-name>After Modify</resource-name>
    </resource-ref>
</root>

task ("replace")<<{
def xmlSource = file(path/to/test.xml)
def xmlDest = file(path/to/destination)
def xmlParser = new XmlParser()
xmlParser.setFeature("http://apache.org/xml/features/disallow-doctype-decl", false)
def xmlRoot = xmlParser.parse(xmlSource)

xmlRoot.'display-name'[0].value = 'RTM16'
//Looking for something like this
//xmlRoot.'resource-ref'[@id='change'].'resource-name'[0].value = 'After Change'
//xmlRoot.'resource-ref'[@id='modify'].'resource-name'[0].value = 'After Modify'

def nodePrinter = new XmlNodePrinter(new PrintWriter(new FileWriter(xmlDest)))
nodePrinter.preserveWhitespace = true
nodePrinter.print(xmlRoot)
}

task（“replace”）在经历了这个过程之后，我想到了这个

task ("replace")<<{
xmlSource = file(path/to/xml source file)
xmlDest = file(path/to/destinationfile)
def parser = new XmlParser()
def xmlRoot = parser.parse(xmlSource)
xmlRoot.each{
    if(it.name().equals("resource-ref")&& it.@id.equals("change")){
        it.'resource-name'[0].value = 'After Change'
    }
    else if(it.name().equals("resource-ref")&& it.@id.equals("modify")){
        it.'resource-name'[0].value = 'After Modify'
    }
}

def b = new XmlNodePrinter(new PrintWriter(new FileWriter(xmlDest)))
b.preserveWhitespace = true
b.print(z)
}

任务（“替换”）