How to packaging specific directory in scala?

Question

Now I have scala project directory like below:

my-project
ᄂsrc/main/scala
   ᄂ com.foo.bar
       ├ foo1
       │  ᄂ (...classes)
       ᄂ foo2

Then I want to package to jar file just 'com.foo.bar.foo1' directory files. But if i use sbt package, sbt make all directories below the src/main/scala.

If I command sbt package:

target
ᄂ some-name-package.jar
    ᄂ com.foo.bar
       ├ foo1
       │  ᄂ (...classes)
       ᄂ foo2

I want to know how to package like below:

target
ᄂ some-specific-package.jar
    ᄂ com.foo.bar
       ├ foo1
          ᄂ (...classes)

But I cannot change build.sbt file.

How can I package specific directory?

Answer 1

You can specify mappings

Compile / packageBin / mappings := {
  val original = (Compile / packageBin / mappings).value
  original.filter {
    case (file, toPath) =>
      toPath.startsWith("com/foo/bar/foo1")
  }
}

Then executing sbt package creates a jar with classes only from com.foo.bar.foo1.*

https://www.scala-sbt.org/1.x/docs/Howto-Package.html#Modify+the+contents+of+the+package

How to create a custom package task to jar a subset of classes in SBT

How to create custom "package" task to jar up only specific package in SBT?

---

Alternatively, you can divide your project into two subprojects. Then the directory structure will be

my-project
ᄂsubproject1
  ᄂsrc/main/scala
     ᄂ com.foo.bar
         ᄂ foo1
            ᄂ (...classes)
ᄂsubproject2
  ᄂsrc/main/scala
     ᄂ com.foo.bar
         ᄂ foo2

build.sbt

lazy val subproject1 = (project in file("subproject1"))
  .dependsOn(...)
  .settings(
    name := "subproject1",
    ...
  )

lazy val subproject2 = (project in file("subproject2"))
  .dependsOn(...)
  .settings(
    name := "subproject2",
    ...
  )

https://www.scala-sbt.org/1.x/docs/Multi-Project.html

It will be enough to execute sbt subproject1/package then.