How do I delete the builds 11 to 1717 in Jenkins?

Question

Due to a misconfiguration, I have over 1700 failed builds in a Jenkins job.

How can I efficiently clean them?

Answer 1

You have several options:

• Temporarily set the number of builds to keep in the job configuration (Discard Old Builds) so that those builds will be deleted when the next build finishes. If the next build is 1800, set it to keep the most recent 85 or so. Mark all older builds (i.e. 1 to 10) as Keep This Build Forever before starting the next build. This option will fail to delete some builds if you have a downstream job that prevents upstream builds from being deleted (not a problem in your situation if all of them failed though).

• Use the Script Console in Manage Jenkins. If it's a top level job (not in a folder), the following will do the trick:

  Jenkins.instance.getItemByFullName('JobName').builds.findAll { it.number > 10 && it.number < 1717 }.each { it.delete() }

  Of course, this approach generally requires good backups. There's a lot you can break using the script console.

• Delete the builds' folders from disk (by default in $JENKINS_HOME/jobs/JobName/builds/, using the start time stamp as folder name) and restart Jenkins, or Reload Configuration From Disk. This option will not allow plugins that e.g. keep the SVN history in order by moving any changes to the subsequent build to do their job.

Answer 2

Simply make an API call:

curl -X POST http://jenkinUser:jenkinAPIToken@yourJenkinsURl.com/job/theJob/[11-1717]/doDelete

To get the APIToken: login to Jenkins > Configuration > Show API Token.

Answer 3

As Aaron stated correctly you can also use the Jenkins CLI for this purpose:

java -jar jenkins-cli.jar -s http://yourserver.com delete-builds <JobName> 11-1717

Answer 4

Template for a script to run in the Jenkins Script-Console. Use the flag reallyDelete to test it before actually deleting:

// Jenkins job
def jobName = 'foo'
// Range of builds to delete
def rs = Fingerprint.RangeSet.fromString("11-1717", false);
// Set to true to actually delete. Use false to test the script.
def reallyDelete = false;

// ----------------------------------
def job = Jenkins.instance.getItemByFullName(jobName);
println("Job: ${job.fullName}");

def builds = job.getBuilds(rs);
println("Found ${builds.size()} builds");
builds.each{ b-> 
  if (reallyDelete) {
    println("Deleting ${b}");
    b.delete();
  } else {
    println("Found match ${b}");
  }
}

Answer 5

I was faced with the same task when I took over a Jenkins server, where there are just over 150 jobs with up to 3000 old builds, so I wrote a small bash script that only keeps the last 10 builds:

#! /bin/bash

initialPath=$(pwd);

find /var/lib/jenkins/ -type d -name builds | while read jobs
 do

    #############################################################
    ## Enter build-directory of current job and get some numbers
    #############################################################
    cd "$jobs" ;
    ls -d [[:digit:]]* &>/dev/null ;
    rc=$? ;
    if [[ $rc -eq 0 ]] ;
     then
        buildName=$(ls -d [[:digit:]]*)  ;
        latestBuild=$(echo $buildName | awk '{print $NF}') ; # highest number
        oldestBuild=$(echo $buildName | awk '{print $1}') ; # lowest number
        amountOfBuilds=$(echo $buildName | wc -w ) ;
        lastBuildToKeep=$(echo "${latestBuild} 9" | awk '{print $1 - $2}') ;

        ############################################################
        ## Skip Folder if it contains less than 10 builds
        ############################################################
        if [ ${amountOfBuilds} -le 10 ] ;
         then
            echo "Skipping $(pwd) --> less than 10 builds";
         else
            ############################################################
            ## Delete all build-directories except the last 10
            ############################################################
            for (( i=$oldestBuild; i<$lastBuildToKeep; i++))
             do
                echo "Deleting $(pwd)/${i} ..."
                rm -r "$i" ;
            done ;
        fi ;
     else
        echo "Skipping $(pwd) --> Zero builds";
    fi ;
done ;

############################################################
## go back to $initialPath
############################################################
cd "$initialPath" ;

Restarting Jenkins afterwards is highly recommended to avoid problems. Thx @Aaron Digulla

Answer 6

Go to Manage Jenkins > Script Console

Run below command

def jobName = "build name"  
def job = Jenkins.instance.getItem(jobName)  
job.getBuilds().each { if(it.number >= 11 && it.number <= 1717) it.delete() }  
job.save()

Answer 7

I have created a small python script which would serve this purpose. Following is the script:

delete_jenkins_builds.py

from os import listdir, path
import shutil


job_build_fullpath = '/var/lib/jenkins/jobs/My-Jenkins-Project/builds'
print listdir(job_build_fullpath)

for build_dir in listdir(job_build_fullpath):
        if build_dir.isdigit() and int(build_dir) in range(11, 1718):
                build_dir_fullpath = path.join(job_build_fullpath, build_dir)
                print "Deleting: " + build_dir_fullpath
                shutil.rmtree(build_dir_fullpath)

• job_build_fullpath - Path of the Job's build directory

• range(start_build_no, end_build_no) - range(11, 1718) looks for all the builds ranging from build no. 11 to build no. 1717. Please set it accordingly.

• shutil.rmtree(build_dir_fullpath) - deletes each build directory that are in the range.

Python Version: 2.7