CODEWITHSUNDEEP
jenkinsgroovyxmlslurper
Ryan Johnson
Ryan Johnson

Reputation: 109

Jenkins Groovy with XMLSlurper Error "failed to serialize"

I am on Jenksin 2.46.2 and have a job that uses Active Choices Reactive Parameter. I select my servers in my first selection from a XML file using XMLSlurper and reference that for my second selection. When I hardcode the server name the code works fine. When I use the variable in my code I get an error.

This code works:

def serverList = new XmlSlurper().parse("/app/jenkins/jobs/servers.xml")
def SERVER = 'testserver1'
def output = []
serverList.Server.find { it.@name == SERVER}.CleanUp.GZIP.File.each{ 
    it.output.add(p)
}

return output

When I reference the variable selection from my previous selection I get the error:

def serverList = new XmlSlurper().parse("/app/jenkins/jobs/servers.xml")
def SERVER = SERVER
def output = []
serverList.Server.find { it.@name == SERVER}.CleanUp.GZIP.File.each{ 
    it.output.add(p)
}

return output

The error that I am getting is below. Any idea why I get an error?

WARNING: failed to serialize [[/app/test2/log], [/app/test2/log]] for ...*other text*... net.sf.json.JSONException: There is a cycle in the hierarchy!

Here is my XML file:

<ServerList>
  <Server name="testserver1">
      <CleanUP>
        <GZIP>
          <File KeepDays="30">/app/test1/log</File>
        </GZIP>
      </CleanUP>
    </Server>  
  <Server name="testserver2">
    <CleanUP>
      <GZIP>
        <File KeepDays="30">/app/test2/log</File>
      </GZIP>
    </CleanUP>
  </Server>
</ServerList>

NE.jpg

Upvotes: 2

Views: 1967

Answers (1)

Rao
Rao

Reputation: 21379

Here is the script that you need, which reads the value of File element and returns a list:

def serverList = new XmlSlurper().parse("/app/jenkins/jobs/servers.xml")
return serverList.'**'.findAll{ it.name() == 'File'}*.text()

Output:

[/app/test1/log, /app/test2/log]

EDIT: based on OP comments

def server = 'testserver1'
def serverList = new XmlSlurper().parse("/app/jenkins/jobs/servers.xml")
def result = serverList.'**'.find{ it.@name == server}.CleanUP.GZIP.File
println result
return result

EDIT2:

If you want list or array, try below:

def server = 'testserver1'
def serverList = new XmlSlurper().parse("/app/jenkins/jobs/servers.xml")
def result = serverList.'**'.findAll{ it.@name == server}*.CleanUP.GZIP.File.text()
println result
return result

Upvotes: 1

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