Groovy XMLSlurper Parse Values - groovy

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

Answer 2

def output = []
def node = serverList.Server.find {
it.'name' = 'testserver1'
}.CleanUp.GZIP.File.each {
output.add(it)
}
return output
also there is a copy & paste error in your .xml. You have to add </ServerList> at the end.
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