Jenkinsfile find files in S3 with s3FindFiles

Question

I wan to use s3FindFiles in Jenkinsfile (pipeline) to search for file in S3 bucket. I tried to use it it following way

        steps {
            withCredentials([[
                $class: 'AmazonWebServicesCredentialsBinding',credentialsId: 'jenkins-user-for-aws',accessKeyVariable: 'AWS_ACCESS_KEY_ID',secretKeyVariable: 'AWS_SECRET_ACCESS_KEY'
            ]]) {
                    s3FindFiles(bucket:'my-bucket', path:'firmwares/', glob:'gwsw_*')
                }
        }

which prints

      Searching s3://my-bucket/firmwares/ for glob:'gwsw_*' 
      Search complete

How do I get name of the files from it ?

As per s3FindFiles it return name, so I tried

    steps {
                withCredentials([[
                    $class: 'AmazonWebServicesCredentialsBinding',credentialsId: 'jenkins-user-for-aws',accessKeyVariable: 'AWS_ACCESS_KEY_ID',secretKeyVariable: 'AWS_SECRET_ACCESS_KEY'
                ]]) {
                        files = s3FindFiles(bucket:'my-bucket', path:'firmwares/', glob:'gwsw_*')
                        echo files[0].name
                    }
            }

But got this error :

 WorkflowScript: 256: Expected a step @ line 256, column 19.
                   files = s3FindFiles(bucket:'my-bucket', path:"firmwares/", glob:'gwsw_*')

Answer 1

s3FindFiles

return an array of FileWrapper instances with the following properties:

• name: the filename portion of the path (for "path/to/my/file.ext", this would be "file.ext")
• path: the full path of the file, relative to the path specified (for path="path/to/", this property of the file "path/to/my/file.ext" would be "my/file.ext")
• directory: true if this is a directory; false otherwise length: the length of the file (this is always "0" for directories)
• lastModified: the last modification timestamp, in milliseconds since the Unix epoch (this is always "0" for directories)

you can simply iterate over the returned array using either the properties above or the dedicated methods

e.g: get file name

name_by_property = files[0].name   
name_by_method = files[0].getName()

if you are using declarative pipeline you'll need to wrap your code with script block:

steps {
        withCredentials([[$class: 'AmazonWebServicesCredentialsBinding', credentialsId: 'jenkins-user-for-aws', accessKeyVariable: 'AWS_ACCESS_KEY_ID', secretKeyVariable: 'AWS_SECRET_ACCESS_KEY']]) {

            script{
              files = s3FindFiles(bucket:'my-bucket', path:'firmwares/', glob:'gwsw_*', onlyFiles: true)
              
              // get the first name
              println files[0].name

              // iterate over all files
              files.each { println "File name: ${it.name}, timestamp: ${it.lastModified}"}
            }
        }
    }