Find Jenkins job through command line

Question

Currently I search with:

ps axf | grep jenkins

And see my job in this format:

51412 ?        S      0:00  \_ /bin/sh -xe /var/lib/jenkins/tmp/jenkins2641742926206351383.sh

The job's name is GOOD_JOB

How can I return jenkins2641742926206351383.sh based on GOOD_JOB only in the command line?

In other words, if I only have GOOD_JOB how do I return jenkins2641742926206351383.sh through command line?

Answer 1

有可能更简单的方法来做到这一点，但你可以用这样的awk做到这一点：

ps axf | grep jenkins | awk -F'/' '{print $8}'

Answer 2

IIUC you want to "Find a Jenkins' build temporary build script name given a Job Name." In the code below I assume you want the last build, or know the build number.

Jenkins has many APIs which might reveal this, but being an old CLI greybeard here's a quick bash recipe to 'grep' by pattern in the Jenkins build logs:

JENKINS_HOME=/path/to/Jenkins/home
find_jenkins_sh() {
    local jobname bldnum log res
    jobname=$1
    bldnum=$2
    ### if we don't get a build number, find the latest one
    if [ -z "$bldnum" ] ; then
         bldnum=$(cd $JENKINS_HOME/jobs/$jobname/builds &&
                  /bin/ls -trd [0-9]* | tail -1)
    fi
    ### now we can locate the log and grep for our pattern
    local log=$JENKINS_HOME/jobs/$jobname/builds/$bldnum/log
    res=$(grep -m1 jenkins[0-9]*.sh $log)   # first mention: -m1
    res=${res//*jenkins/jenkins}            # strip up to 'jenkins'
    echo "$log: $res"
}

# if you know the job number
find_jenkins_sh GOOD_JOB 1234

# use the last job number
find_jenkins_sh GOOD_JOB