Jenkins参数化构建不创建文件

Question

Maybe I am misunderstanding the intended use for the Jenkins file parameter here...

I want to be able to upload a file containing some data (in my case comma separated variables). I then want to simply read this file and do stuff with the data. I've got this setup using a Pipeline job.

My file location is set to 'email_list.csv'. In my pipeline script I have

node {
  stage('post') {
    emailFile = readFile 'email_list.csv'
    println "${emailFile}"
    //.........
  }
}

This fails with a java.io.FileNotFoundException: /var/lib/jenkins/workspace/job-name/email_list.csv (No such file or directory) exception

Shouldn't the parameterized build have set up this file? If not, how do I read the data uploaded?

Answer 1

Jenkins by default provides build job parameters as a params map in pipeline. It is a key-value pair. All you comma seperated values will be into values field. You can refer them in your groovy script as,

print params.emailFile

To dump it as a file, you can use writeFile library function.

PS: If you print params in groovy script, you will be able to see all build parameters of your job.

Answer 2

There is aa bug since ages that makes impossible to use fileParameter :

• Handle file parameters
• file parameter not working in pipeline job