0

我有以下 gradle 项目结构：

SUPER/
|______A/
|       |______A1/
|       |       |______build.gradle
|       |______A2/
|       |       |______build.gradle
|       |______build.gradle
|       |______settings.gradle
|
|______B/
|       |______build.gradle
|       |______settings.gradle
|
|______build.gradle
|______settings.gradle

我的 A-Project 中的 build.gradle包含以下内容：

...
project(":A1") {
    apply plugin: "java-library"


    dependencies {
        api project(":A2")
    }
}

project(":A2") {
    apply plugin: "java-library"


    dependencies {
        ...
    }
}
...

我的“Super”-settings.gradle文件包含类似的内容：

rootProject.name = 'SUPER'

include 'A'
include 'B'

在我的子项目“A”中，settings.gradle 看起来像这样：

include 'A1', 'A2'

当从 SUPER 文件夹在命令行上像“ ./gradlew SUPER:A1:sometask”那样调用 gradlewrapper 时，它会失败并显示以下消息：

FAILURE: Build failed with an exception.

* Where:
Build file 'C:\Users\<somepath>\SUPER\A\build.gradle' line: 44

* What went wrong:
A problem occurred evaluating project ':A'.
> Project with path ':A1' could not be found in project ':A'.

./gradlew A1:sometask但是，当从 A-sub-project-folder 中执行类似“ ”的操作时，它可以成功运行。

你能解释一下我在这里做错了什么吗？

4

0 回答 0