scala - SBT项目导航

Question

我有以下版本:

  lazy val stampleSearchProject = RootProject(file("../stample-search"))


  lazy val main = play.Project(appName, appVersion, appDependencies)
    .dependsOn(stampleSearchProject)
    .settings(defaultScalaSettings:_*)
    .settings( ......... )

戳搜索项目有一个 build.sbt 文件，其中包含 name := "stample-search"
我在 SBT 文档中找到了:

Navigating projects interactively: At the sbt interactive prompt, type projects to list your projects and project to select a current project. When you run a task like compile, it runs on the current project. So you don't necessarily have to compile the root project, you could compile only a subproject.

但是 SBT 文档不使用 RootProject 而是使用 Project (这似乎需要在根项目中使用项目路径，我的意思是我不能使用 ../my-project )

所以我尝试了这个。

[Stample] $ projects
[info] In file:/home/sebastien/Bureau/Stample/stample-web/
[info]   * Stample
[info] In file:/home/sebastien/Bureau/Stample/stample-search/
[info]     default-ccdbaa

所以它似乎列出了两个项目，但我想知道这个名字是从哪里来的 default-ccdbaa
然后我尝试使用项目命令

[Stample] $ project Stample
[info] Set current project to Stample (in build file:/home/sebastien/Bureau/Stample/stample-web/)

[Stample] $ project default-ccdbaa
[error] Invalid project name 'default-ccdbaa' in build file:/home/sebastien/Bureau/Stample/stample-web/ (type 'projects' to list available projects).

[Stample] $ project stample-search
[error] Invalid project name 'stample-search' in build file:/home/sebastien/Bureau/Stample/stample-web/ (type 'projects' to list available projects).

它似乎不起作用。有人可以解释我的构建有什么问题吗？

Answer 1

有关详细信息，您可能需要阅读 Full Configuration (Draft) : Project Relationships

There are three kinds of project relationships in sbt. These are described by execution, classpath, and configuration dependencies. ...

RootProject定义项目引用，而不是子项目。所以你有效地创建了一个源级依赖，但你不能切换到它。更好的方法是使用普通 Multi-Project build .

import sbt._
import Keys._

object HelloBuild extends Build {
  lazy val root = Project(id = "hello",
                          base = file(".")) aggregate(foo, bar)

  lazy val foo = Project(id = "hello-foo",
                         base = file("foo"))

  lazy val bar = Project(id = "hello-bar",
                         base = file("bar"))
}