Jenkins 因 grep 结果为空而失败

Question

我需要在 jenkins 中过滤一个文件。只要结果不为空，过滤就会起作用。但是，如果结果输出为空，管道将失败并显示 ERROR: script returned exit code 1 Finished: FAILURE

示例：

#!groovy
pipeline {
    agent any
    
    stages {

        stage ('mystage') {

            steps {
                    script {

                        sh "echo '' > myfile"
                        sh "echo 'foo 0' >> myfile"
                        sh "echo 'foo 1' >> myfile"
                        
                        sh "grep foo myfile"
                        sh "grep ba myfile"
                    }
            }
        }        
    }
}

输出：

+ echo ''
[Pipeline] sh
+ echo 'foo 0'
[Pipeline] sh
+ echo 'foo 1'
[Pipeline] sh
+ grep foo myfile
foo 0
foo 1
[Pipeline] sh
+ grep ba myfile
[Pipeline] }
[Pipeline] // script
[Pipeline] }
[Pipeline] // stage
[Pipeline] }
[Pipeline] // withEnv
[Pipeline] }
[Pipeline] // node
[Pipeline] End of Pipeline
ERROR: script returned exit code 1
Finished: FAILURE

使用 grep ba myfile > catchoutput 将输出路由到文件不起作用。我怎样才能输出 grep 结果，而管道在这种边缘情况下不会失败？

添加像 sh "echo 'dummyline that won't match' >> myfile" 这样的虚拟行似乎可行，但却是一种 hack。有没有干净的解决方案？

Answer 1

我们可以在变量中取 return 值：

def ret = sh(script: 'grep ba myfile', returnStdout: true)

更多信息：https://www.jenkins.io/doc/pipeline/steps/workflow-durable-task-step/#code-sh-code-shell-script。
注意你也可以加上returnStatus: true，这样jenkins step即使出现命令失败也不会失败

Jenkins 因 grep 结果为空而失败

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