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，此依赖关系定义如下：

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

而且，project_3依赖于project_2，此依赖关系定义如下：

- 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，会发生以下错误：

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

我希望能够从其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

然后从所有子项目的project(":project_#").projectDir = new File("../project_#")文件中删除settings.gradle行。但是即使如此，每个项目目录中的build命令也无法使用。

Answer 1

请勿在模块级include(":project_1/2/3")中定义build gradle，而应使用根项目的settings.gradle来定义此内容...

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

，然后您可以在模块级build.gradle中引用这些子项目：

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

Answer 2

项目依赖项需要一个项目路径，而不仅仅是项目名称：

dependencies {
    implementation project(":project_1")
}

在项目名称前注意:：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，我有一些不是 Java 项目的子项目。我必须这样做

allprojects {
   apply plugin: 'java'
}

在 build.gradle 中

gradle-项目作为依赖项

3 个答案: