I know that Jenkinsfile pipeline DSL can resolve libraryResource()
magically without explicitly providing any context (pardon my poor
understanding of the pipeline DSL).
But if I want to use libraryResource() as a regular groovy class, how do I
use it???
Is it a static method? Which class does
When invoked from Jenkinsfile, pipeline steps work fine and all. But when I
want to compile a library with pipeline steps, compilation fails.
>
> package com.company;
>
> def read(fileName) {
> return libraryResource(fileName)
> }
>
> return this
>
>
How to compile the above code as regular .gr
Hello,
package com.mycompany;
> import org.jenkinsci.plugins.workflow.libs.*;
>
> this.metaClass.mixin(ResourceStep)
> def read(fileName) {
> return libraryResource(fileName)
> }
>
> return this
>
>
I have the above LibraryResource.groovy file that basically just delegates
to the libraryResourc
Thank you.
On Friday, 2 December 2016 01:08:40 UTC-5, Baptiste Mathus wrote:
>
> Because that file always changes. It's like 'apt-get update'. You want to
> run it to see an up-to-date list of plugins.
>
> Cheers
>
> Le 1 déc. 2016 10:51 PM, "TInaTinaTian
Hello,
I came across this script that updates jenkins UpdateCenter from command
line: https://gist.github.com/jedi4ever/898114
> # Get the update center ourself
>
> $ wget -O default.json http://updates.jenkins-ci.org/update-center.json
>
> # Now push it to the update URL
>
> curl -X POST -H "A
