Reputation: 1419
I am in the process of cleaning up Jenkins (it was setup incorrectly) and I need to delete builds that are older than the latest 20 builds for every job.
Is there any way to automate this using a script or something?
I found many solutions to delete certain builds for specific jobs, but I can't seem to find anything for all jobs at once.
Any help is much appreciated.
Upvotes: 18
Views: 31587
Reputation: 476
I had issues running the suggestions on my Jenkins instance. It could be because it is dockerized. In any case, removing the folder beforehand using the underlying bash interpreter fixes the issue. I also modified the script to keep 180 days of build logs and keep a minimum of 7 build logs:
import jenkins.model.Jenkins
import hudson.model.Job
MIN_BUILD_LOGS = 7
def sixMonthsAgo = new Date() - 180
Jenkins.instance.getAllItems(Job.class).each { job ->
println job.getFullDisplayName()
def recent = job.builds.limit(MIN_BUILD_LOGS)
def buildsToDelete = job.builds.findAll {
!recent.contains(it) && ! (it.getTime() > sixMonthsAgo)
}
if (!buildsToDelete) {
println "nothing to do"
}
for (build in buildsToDelete) {
println "Preparing to delete: " + build + build.getTime()
["bash", "-c", "rm -r " + build.getRootDir()].execute()
build.delete()
}
}
"done"
Upvotes: 0
Reputation: 15972
You can use the Jenkins Script Console to iterate through all jobs, get a list of the N most recent and perform some action on the others.
import jenkins.model.Jenkins
import hudson.model.Job
MAX_BUILDS = 20
for (job in Jenkins.instance.items) {
println job.name
def recent = job.builds.limit(MAX_BUILDS)
for (build in job.builds) {
if (!recent.contains(build)) {
println "Preparing to delete: " + build
// build.delete()
}
}
}
The Jenkins Script Console is a great tool for administrative maintenance like this and there's often an existing script that does something similar to what you want.
Upvotes: 38
Reputation: 3330
I got an issue No such property: builds for class: com.cloudbees.hudson.plugins.folder.Folder
on Folders Plugin 6.6 while running @Dave Bacher's script
Alter it to use functional api
import jenkins.model.Jenkins
import hudson.model.Job
MAX_BUILDS = 5
Jenkins.instance.getAllItems(Job.class).each { job ->
println job.name
def recent = job.builds.limit(MAX_BUILDS)
for (build in job.builds) {
if (!recent.contains(build)) {
println "Preparing to delete: " + build
build.delete()
}
}
}
Upvotes: 14
Reputation: 183
For Multibranch Pipelines, I modified the script by Dave Bacher a bit. Use this to delete builds older than the latest 20 build of "master" branches:
MAX_BUILDS = 20
for (job in Jenkins.instance.items) {
if(job instanceof jenkins.branch.MultiBranchProject) {
job = job.getJob("master")
def recent = job.builds.limit(MAX_BUILDS)
for (build in job.builds) {
if (!recent.contains(build)) {
println "Preparing to delete: " + build
// build.delete()
}
}
}
}
Upvotes: 3
Reputation: 9075
There are lots of ways to do this
Personally I would use the 'discard old builds' in the job config
If you have lots of jobs you could use the CLI to step through all the jobs to add it
Alternatively there is the configuration slicing plugin which will also do this for you on a large scale
Upvotes: 3
Reputation: 181
This can be done in many ways. You can try the following
ls >jobs.txt
Now you can write a shell script with a for loop
#!/bin/bash
##read the jobs.txt
for i in 'cat <pathtojobs.txt>'
do
curl -X POST http://jenkins-host.tld:8080/jenkins/job/$i/[1-9]*/doDeleteAll
done
the above deletes all the jobs
you can also refer here for more answers
Upvotes: 1