Java 如何通过 Groovy 获取文件 build.gradle 的完整路径？
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How get full path to file build.gradle by Groovy?
提问by
I need create file in my java-project near buiid.gradle-file. I must create task (Groovy task) in build.gradle-file, my task must create file near buiid.gradle in my project, but I do not know - how do get path to buiid.gradle-file, that is putting in my project.
我需要在靠近 buiid.gradle 文件的 java 项目中创建文件。我必须在 build.gradle-file 中创建任务（Groovy 任务），我的任务必须在我的项目中的 buiid.gradle 附近创建文件，但我不知道 - 如何获取 buiid.gradle-file 的路径，这是放入我的项目。
How get full path to file buiid.gradle by Groovy? Help me, please.
如何通过 Groovy 获取文件 buiid.gradle 的完整路径？请帮帮我。
采纳答案by kuporific
There are several ways to accomplish this.
有几种方法可以实现这一点。
- If you look at the Working With Filespage, you can simply use the
file()method that is a part of theProjectobject. - If you look at the Project DSL Docs, you should see the
projectDirproperty. Thes properties are available throughout thebuild.gradlescript.
- 如果您查看“使用文件”页面，您可以简单地使用
file()作为Project对象一部分的方法。 - 如果您查看Project DSL Docs，您应该会看到该
projectDir属性。这些属性在整个build.gradle脚本中都可用。
They can be used, respectively, like this:
它们可以分别使用，如下所示：
task myTask << {
println file('.')
println projectDir
}
Would output
会输出
/full/path/to/your/project/ /full/path/to/your/project/
Where that directory contains the build.gradlefile where the task is located.
该目录包含build.gradle任务所在的文件。
回答by Stephen Talley
To get a handle to the java.io.File that represents the currently-executing .gradle file:
要获取代表当前正在执行的 .gradle 文件的 java.io.File 的句柄：
def file = buildscript.sourceFile
In my case, I needed the directory of that script so I could apply other configuration from the same directory:
就我而言，我需要该脚本的目录，以便我可以从同一目录应用其他配置：
def buildCommonDir = buildscript.sourceFile.getParent()
apply from: "$buildCommonDir/common.gradle"
This approach works with any.gradle file, not just build.gradle.
这种方法适用于任何.gradle 文件，而不仅仅是 build.gradle。
