7

I am one short step away from being able to replace numerous customized Jenkinsfiles with one. All I need to get is the directory containing the Jenkinsfile under execution. I am using the Declarative pipeline syntax.

I looked at several ideas in groovy code, but could not find how to obtain this.

For those that need to see something specific, a Jenkinsfile looks like this at the top:

#!/usr/bin/env groovy pipeline { agent any environment { tf='/var/lib/jenkins/tools/terraform0108/terraform' dir='clients/xyz/us-east-1/dev' }

That variable dir is assigned to the location of that Jenkinsfile. dir is accessed later in the groovy code. I want to be able to omit that assignment and just pick up that directory from context of the executing script.

I tried several different things like steps containing

script { println __FILE__ }

and

script { scriptDir = new File(getClass().protectionDomain.codeSource.location.path).parent println scriptDir }

Neither of which ran (gave non-existent variable in FILE case and permission violation in the second case). I tried "${FILE}" and other variants.

I need to use the directory in a steps -- sh block so I believe it needs to be in an environment item.

Now, the Jenkins job configuration gives the path to the Jenkins file, but I don't want to have to repeat that to create another environment variable at that level.

Already consulted:

Help is appreciated.

Improve this question

3 Answers 3

Reset to default
2

There are lots of ways to do this. Here are two ways I can think of off the top of my head:

steps { println(WORKSPACE) }

or

steps { def foo = sh(script: 'pwd', returnStdout: true) println(foo) }
Improve this answer
1
  • 5
    Ah, I bet you are assuming my Jenkinsfile will be at the top of the workspace. It is located down into subdirectories. Using WORKSPACE or pwd only give the Workspace directory for the job, not the location of the Jenkinsfile it is executing. Commented Aug 15, 2018 at 14:18
1

In my MultiBranchPipeline I've achieved the goal using this shared library code:

#!groovy import jenkins.model.Jenkins String call() { String thisMultiBranchProjectName = JOB_NAME.split('/')[0] def thisMultiBranchProject = Jenkins.getInstance().getItemByFullName(thisMultiBranchProjectName) def thisBranchProjectFactory = thisMultiBranchProject.getProjectFactory() String thisStringPath = thisBranchProjectFactory.getScriptPath() return thisStringPath }

I do concede that this looks more like a hack…

Improve this answer
0

You're asking about pipelines, not JobDSL, but hopefully this helps/applies:

In my DSL seed jobs I use:

// Boilerplate def dslWorkspacePath = new File(__FILE__).parent evaluate(new File("${dslWorkspacePath}/scaffolding.groovy"))

in order to determine the path of the seed job that is currently being processed by JobDSL and then run some setup code.

My seed jobs do not sit in the root of the repo that gets cloned, but rather something like /src/main/groovy/jobdsl/

Improve this answer
3
  • Thanks for trying! I can easily get the Workspace directory. However, Pipeline allows you to reference a Jenkinsfile with a path (it is not at the top of the workspace) in the setup. I need that path. Commented Feb 11, 2019 at 16:49
  • Sorry, should have checked that. __FILE__ returns groovy.lang.MissingPropertyException: No such property: __FILE__ for class. Keep in mind that what you see in the config is where it's been configured to pull from SCM, not where it lives once it's been pulled. That's usually something like jobname@script folder. (I'm sure you know this already) Commented Feb 12, 2019 at 9:54
  • 1
    Also, consider that there may not always be a path to the running script (ie, when it's entered directly into the job config, not pulled from scm) Commented Feb 12, 2019 at 10:09

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.