Android Gradle read flavour file

Question

My app has three flavours (flavourOne, flavourTwo and flavourThree) and each flavour has its own versions.properties

I need, at build time, to retrieve a value in that file.

If I do this:

defaultConfig {
    ....
    def file = file("${project.projectDir}/src/flavourOne/version.properties")
    // I can now access to the file and obtain the value that I want.
}

But in the above solution, I need to explicitly set the path to the file. Since I want the file that is contained I would like to do something like:

def file = file("${project.projectDir}/src/../version.properties") and dependending of the selected flavour that file would be transformed into:

def file = file("${project.projectDir}/src/flavourOne/version.properties")
def file = file("${project.projectDir}/src/flavourTwo/version.properties")
def file = file("${project.projectDir}/src/flavourThree/version.properties")

Instead, what the above does its to set the file to ${project.projectDir}/version.properties

How can I read the specific flavour file at build time?

I think this might help : https://stackoverflow.com/a/30621515/2199001 — Chris, Oct 18 '21 at 16:33
@Chris Thanks. Within that question I've found a way to achieve what I needed. — Favolas, Oct 20 '21 at 08:46
Cool. Glad I could help. You could now post a solution to help others that come across your question. — Chris, Oct 21 '21 at 05:38

Android Gradle read flavour file

0 Answers0