Groovy XMLSlurper Parse Values

Question

I am so close to getting my code completed. I would like to get only the values in an array. Right now I am getting XML declaration plus the line.

Here's my code:

import groovy.xml.XmlUtil

def serverList = new 
XmlSlurper().parse("/app/jenkins/jobs/firstsos_servers.xml")

def output = []
serverList.Server.find { it.@name == SERVER}.CleanUp.GZIP.File.each{
     output.add(XmlUtil.serialize(it))
}

return output

Here is my XML File:

<ServerList>
    <Server name="testserver1">
            <CleanUp>
                    <GZIP>
                            <File KeepDays="30">log1</File>
                            <File KeepDays="30">log1.2</File>
                    </GZIP>
            </CleanUp>
    </Server>
    <Server name="testserver2">
            <CleanUp>
                    <GZIP>
                            <File KeepDays="30">log2</File>
                    </GZIP>
            </CleanUp>
    </Server>
    <Server name="testserver3">
            <CleanUp>
                    <GZIP>
                            <File KeepDays="30">log3</File>
                    </GZIP>
            </CleanUp>
    </Server>

When I select testserver1 my output should be:

['log1','log1.2']

What I am getting is this:

<?xml version="1.0" encoding="UTF-8"?><File KeepDays="30">log1</File>
<?xml version="1.0" encoding="UTF-8"?><File KeepDays="30">log2</File>

Answer 1

You need not require to use XmlUtil.serialize()

Here is what you need and following inline comments.

//Define which server you need
def SERVER = 'testserver1'
//Pass the 
def serverList = new 
XmlSlurper().parse("/app/jenkins/jobs/firstsos_servers.xml")

//Get the filtered file names
def output = serverList.Server.findAll{it.@name == SERVER}.'**'.findAll{it.name() == 'File'}*.text()

println output
return output

Output:

You can quickly try online Demo