Using AzureDevOps pipeline's variables in bash Script

alangumanav 0
Viewed 35

My question is quite simple, I'm trying to learn AzureDevOps. I have a pipeline. In this pipeline I have a task with a bash script. This task basically adds files to the archive. This archive format:

  I want it to be MyPackage_09192022_MyDeploymentComment.zip

For this, I created a variable called DeploymentComment in the pipeline. When I start a queque from this pipeline, I fill in the DeploymentComment field. I added this bash script to the task as filepath, so it gives the address of the file on the machine. I also gave the $DeploymentComment variable to the arguments in this task.

My script is as follows

  date="$(date +"%d%m%Y")"
  zipName="MyPackage_"$date"_"$1
  zip -r $zipname /home/admins/myDir/*

I am waiting for the content of the $DeploymentComment variable that I gave as an argument on the Pipeline to come to the part I specified as $1 in the script. In other words, when I start the queque, when I type my1stTry in the $DeploymentComment section, I expect the zip file created when I type my1stTry

     I expect it to be MyPackage_09192022_my1stTry.zip but bash does not see this variable.
     I can create a zip file as MyPackage_09192022_.zip.

What am I missing, can you help me?

azure-pipelines shell bash azure-devops
1 Answers

This will achieve your requirements:

trigger:
- none

pool:
  vmImage: ubuntu-latest
variables:
- name: DeploymentComment #Define the variable
  value: MyDeploymentComment
steps:
- bash: |
    xxx="$(DeploymentComment)" #Use the pipeline variable
    date="$(date +"%d%m%Y")"
    zipName="$(System.DefaultWorkingDirectory)/MyPackage_"$date"_"$xxx
    echo $zipName
    zip -r $zipName.zip ./*
    ls

Successfully get the value on my side:

enter image description here

Bowman Zhu-MSFT0
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