如何在不知道文件名(或编号)的情况下从共享库的资源文件夹中加载文件? [英] How to load files from resources folder in Shared library without knowing their names (or number)?

问题描述

您知道，在Jenkins的共享库中，可以通过执行以下操作来加载资源(位于资源文件夹中):

As you know, in Shared libraries in Jenkins, it is possible to load a resource (located in resources folder) by doing:

libraryResource("script.sh")

现在，我的用例是我想在资源下的文件夹中加载文件数量:

Now my use case is that I want to load number of files inside a folder under resources :

+ resources
 + teamA
  + script1.sh
  + script2.sh

我想在做任何事情之前先加载所有这些文件: 我在共享库中做了一个方法:

And I want to load all those files before doing anything : I did a method in the shared library:

new File(scriptsFolder).eachFile() { file->

 writeFile([file:"${env.workspace}/${file.getName()}",text:libraryResource("$scriptsFolder/${file.getName()}")])
 sh("chmod +x ${env.workspace}/${file.getName()}")
}

其中scriptsFolder= "teamA"

当然，我会得到java.io.IOException: Is a directory 因为libraryResource必须获取文件路径参数.

Of cource I'm getting java.io.IOException: Is a directory Because libraryResource must get a file path parameter.

那么，有什么方法可以加载所有这些文件而无需知道它们的名称或编号吗?

So is, there a way to load all those files without knowing their names or their number?

推荐答案

您可以使用Groovy @SourceURI批注获取共享库的绝对路径:

You can get absolute path of your shared library using Groovy @SourceURI annotation:

// vars/get_resource_dir.groovy
import groovy.transform.SourceURI
import java.nio.file.Path
import java.nio.file.Paths

class ScriptSourceUri {
    @SourceURI
    static URI uri
}

def call() {
    Path scriptLocation = Paths.get(ScriptSourceUri.uri)
    return scriptLocation.getParent().getParent().resolve('resources').toString()
}

使用该路径，您可以照常调用脚本:

Using the path you can invoke your scripts as usual:

sh "${get_resource_dir()}/com/example/test.sh"

这篇关于如何在不知道文件名(或编号)的情况下从共享库的资源文件夹中加载文件?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！