gradle - 项目作为依赖项

Question

我的Java项目结构如下：

- root-directory (has no .gradle file)
    - project_1
        - build.gradle
        - settings.gradle
    - project_2
        - build.gradle
        - settings.gradle
    - project_3
        - build.gradle
        - settings.gradle

project_2

依赖于

project_1

，这种依赖关系定义如下：

- project_2/build.gradle
  implementation(project(":project_1"))
- project_2/settings.gradle
  include(":project_1")
  project(":project_1").projectDir = new File("../project_1")

并且，

project_3

依赖于

project_2

，这种依赖关系定义如下：

- project_3/build.gradle
  implementation(project(":project_2"))
- project_3/settings.gradle
  include(":project_2")
  project(":project_2").projectDir = new File("../project_2")

现在，当我通过在

project_3

目录中运行

>> ./gradlew build

终端命令来构建

./project_3/

时，会出现以下错误：

A problem occurred evaluating project ':project_2'.
> Project with path ':project_1' could not be found in project ':project_2'.

我希望能够从它自己的目录构建每个项目。我该如何解决这个问题？

更新：我尝试通过添加以下内容将

root-directory

转换为项目根目录：

- root-directory/settings.gradle
  rootProject.name = "root"
  include(":project_1", ":project_2", ":project_3")
- root-directory/build.gradle
  // Nothing in this file

并从所有子项目的

project(":project_#").projectDir = new File("../project_#")

文件中删除了

settings.gradle

行。但即使在此之后，每个单独项目目录中的构建命令也不起作用。

Answer 1

不要在模块级别

include(":project_1/2/3")

中定义

build gradle

，而是使用根项目的

settings.gradle

将它们定义为模块：

rootProject.name = "SomeApp"
include ":project_1", ":project_2", ":project_3"

然后就可以在模块级别引用这些子项目了

build.gradle

:

dependencies {
     api project(':project_1')
     implementation project(':project_2')
}

Answer 2

项目依赖项需要项目路径，而不仅仅是项目名称：

dependencies {
    implementation project(":project_1")
}

注意项目名称前的

:

：它将项目名称转换为相对于根项目的项目路径。

Answer 3

就我而言，我的子项目不是java项目。我必须这样做

allprojects {
   apply plugin: 'java'
}

在 build.gradle 中