As you know, in Shared libraries in Jenkins, it is possible to load a resource (located in resources folder) by doing:
libraryResource("script.sh")
Now my use case is that I want to load number of files inside a folder under resources :
+ resources
+ teamA
+ script1.sh
+ script2.sh
And I want to load all those files before doing anything : I did a method in the shared library:
new File(scriptsFolder).eachFile() { file->
writeFile([file:"${env.workspace}/${file.getName()}",text:libraryResource("$scriptsFolder/${file.getName()}")])
sh("chmod +x ${env.workspace}/${file.getName()}")
}
where scriptsFolder= "teamA"
Of cource I'm getting java.io.IOException: Is a directory
Because libraryResource
must get a file path parameter.
So is, there a way to load all those files without knowing their names or their number?
You can get absolute path of your shared library using Groovy @SourceURI annotation:
// vars/get_resource_dir.groovy
import groovy.transform.SourceURI
import java.nio.file.Path
import java.nio.file.Paths
class ScriptSourceUri {
@SourceURI
static URI uri
}
def call() {
Path scriptLocation = Paths.get(ScriptSourceUri.uri)
return scriptLocation.getParent().getParent().resolve('resources').toString()
}
Using the path you can invoke your scripts as usual:
sh "${get_resource_dir()}/com/example/test.sh"
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