如何通过管道忽略作业是否存在来构建Jenkins作业? [英] How to build a Jenkins job from pipeline ignoring if the job doesn't exist?

问题描述

我正试图从我的Jenkins管道中建立工作，像这样:

I'm trying to build a job from my Jenkins pipeline like this:

 build job:"${jobName}", propagate:false, wait:false

这里$ {jobName}是一个多分支管道作业，因此有时在我的工作流程中可能不存在.

Here ${jobName} is a multi branch pipeline job and as such may sometimes not exist in my workflow.

如果作业不存在，此步骤会将我的构建标记为失败. 如果工作不存在，是否有办法简单地忽略并继续前进?

This step marks my build as failed if the job doesn't exist. Is there a way to simply ignore and move on if the job doesn't exist?

我试图像这样检查给定的工作是否存在:

I tried to check whether the given job exist or not like this:

    if(jenkins.model.Jenkins.instance.getItem("${jobName}") != null) {
        println("Preparing to build the ${jobName}...")
        build job:"${jobName}", propagate:false, wait:false
    } else {
        println("Not building the job ${jobName} as it doesn't exist")
    }

但是，有时这会失败(由于脚本安全?). 有更好的方法吗? 我需要做的就是仅在存在的情况下建立工作.

However, this at times fails (due to script security?). Is there a better way of doing this? All I need is to build a job only if it exists.

推荐答案

如果使用脚本化管道，则可以添加try-catch块:

If you use a scripted pipeline, you can add a try-catch block:

try {
    println("Preparing to build the ${jobName}...")
    build job:"${jobName}", propagate:false, wait:false
} catch (NullPointerException e) {
    println("Not building the job ${jobName} as it doesn't exist")
}

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