How to create an instance of org.gradle.api.file.Director

Question

I want to create an instance of org.gradle.api.file.Directory. From the Gradle docs I see that the only way to do this is to create the first instance using project.getLayout().getProjectDirectory() and then use the instance method dir(<path>) on this instance to create an instance for another directory.

Is there a way to directly create an instance of Directory class (like using a File object or directly using a string path)?

Answer 1

When I started working with the new Gradle Lazy Configuration API, I encountered the same problem. Even though dir(<path>) allows absolute paths and you may therefore construct a Directory instance for any directory, it appeared to me like bad design.

However, actually it is a pretty consistent design, because using the old API we just called Project.file(...) on our paths, which evaluated relative paths based on the project directory, too. Constructing File instances directly using its constructor was always a bad idea using Gradle. We may now just replace the calls to file(<path>) with calls to layout.projectDirectory.dir(<path>) or .file(<path) and get the same behavior.

Starting from version 6.0, Gradle also provides a method dir(Provider<File>) via its ProjectLayout class. You may construct the Provider<File> using the method provider(Callable<>) of the Project class, but whether this is actually useful depends on your specific use case.