Shortest way to get File object from resource in Groovy

Question

Right now I'm using Java API to create file object from resource:

new File(getClass().getResource('/resource.xml').toURI())

Is there any more idiomatic/shorter way to do that in Groovy using GDK?

Answer 1

Depending on what you want to do with the File, there might be a shorter way. Note that URL has GDK methods getText(), eachLine{}, and so on.

Illustration 1:

def file = new File(getClass().getResource('/resource.xml').toURI())
def list1 = []
file.eachLine { list1 << it }

// Groovier:
def list2 = []
getClass().getResource('/resource.xml').eachLine {
    list2 << it
}

assert list1 == list2

Illustration 2:

import groovy.xml.*
def xmlSlurper = new XmlSlurper()
def url = getClass().getResource('/resource.xml')

// OP style
def file = new File(url.toURI())
def root1 = xmlSlurper.parseText(file.text)

// Groovier:
def root2 = xmlSlurper.parseText(url.text)

assert root1.text() == root2.text()