I have an sbt build with 2 duplicated projects configuration. See example:
lazy val MyProjectOne = Project(id = "OneId", base = file("path/OneId"))
.dependsOn(moduleOne)
.settings(plugin.settings: _*)
.settings(defaultSettings: _*)
.settings(webSettings: _*)
.settings(libraryDependencies ++= commonTests)
lazy val MyProjectTwo = Project(id = "TwoId", base = file("path/TwoId"))
.dependsOn(moduleOne)
.settings(plugin.settings: _*)
.settings(defaultSettings: _*)
.settings(webSettings: _*)
.settings(libraryDependencies ++= commonTests)
It is obvious that MyProjectOne
and MyProjectTwo
differs only in id
and base
properties.
Is there a way to refactor sbt build like this:
lazy val template = Project()
.dependsOn(moduleOne)
.settings(plugin.settings: _*)
.settings(defaultSettings: _*)
.settings(webSettings: _*)
.settings(libraryDependencies ++= commonTests)
//Just as example:
lazy val MyProjectOne = Project(id = "OneId", base = file("path/OneId")).extends(template)
lazy val MyProjectTwo = Project(id = "TwoId", base = file("path/TwoId")).extends(template)
How can I do that with sbt?
Also
With maven I can define a parent project pom for that case. Is there analog in sbt?