How to get a specific node name in an XML using XPath (using Groovy)

Question

I have an XML as below:

String xml = """  
    <?xml version="1.0" encoding="UTF-8"?>
    <soapenv:Envelope>
        <soapenv:Body>
            <processAsyncRequest></processAsyncRequest>
        </soapenv:Body>
    </soapenv:Envelope>  
"""

Using Groovy, i want to find the first node name inside "Body" using XPath (//:Body/[1]/name())

Output Should be : processAsyncRequest

I am parsing the xml using XmlSlurper. But couldnt figure out the next steps to get node name using XPath

def rootNode = new XmlSlurper().parseText(xml)

Is there any simple way to achieve this? Note : the structure of my xml changes every time. So i want to use Relative Path.

daggett · Accepted Answer · 2018-06-13 21:38:39Z

1

but that's not XPath. it's GPath

String xml = """<?xml version="1.0" encoding="UTF-8"?>
    <soapenv:Envelope xmlns:soapenv='...'>
        <soapenv:Body>
            <processAsyncRequest></processAsyncRequest>
        </soapenv:Body>
    </soapenv:Envelope>  
"""  
def rootNode = new XmlSlurper().parseText(xml)

println rootNode.Body.'*'[0].name()

answered Jun 13, 2018 at 21:38

daggett

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