[SOLVED] Get the Application Package name in template Wizard recipe class Intellij

Get the Application Package name in template Wizard recipe class Intellij

Get the Application Package name in template recipe class

I am working on MVVM plugin template for android studio. I completed most of the work.
But now I only need to get the application package name.
Their is a method

getPackageName(module: Module)
This method takes module as a input But I don't know how to pass the module as input.
My core requiremnet for this task is I want to create a directory in parent directory by stand in child directory.

For example:

I have four directories A, B, C, D.

A(parent) , B ( Child of A) , C (child of B) , D (Child of C).

let say I am standing in directory D and I want to create common directory (F) inside A along with B ( F sibling to B).

Solution

You can get the parent directory form srcOut like

fun getApplicationPackageFile(srcOut: File, applicationPackage: String): File {
    var applicationPackageFile = srcOut.path.toString()
    var pk = applicationPackage.replace(".", "\\")

    val status: Boolean = applicationPackageFile.contains(pk)
    return if (status) {
        var file =
            applicationPackageFile.substring(0, applicationPackageFile.indexOf(pk)) + pk + "\\"
        File(file)
    } else {
        srcOut
    }
}

And call it like

val pkFile = getApplicationPackageFile(srcOut, moduleData.projectTemplateData.applicationPackage)

So If you want to save some file to the core application package..

pkFile.resolve("AppViewModel.$ktOrJavaExt")

For Linux and Mac Change above line to

var pk = applicationPackage.replace(".", "/")