Question

我确定这是微不足道的，但我找不到解决方法...

在我的build.gradle中，我希望processResources任务创建（而不是复制或填充某些模板）要由Java程序加载的资源文件。

我实现了以下目标：

processResources {
    ...

    // This is a collection of files I want to copy into resources.
    def extra = configurations.extra.filter { file -> file.isFile () }

    // This actually copies them to 'classes/extra'. It works.
    into ('extra') {
        from extra
    }

    doLast {
        // I want to write this string (list of filenames, one per
        // line) to 'classes/extra/list.txt'.
        println extra.files.collect { file -> file.name }.join ("\n")
    }
}

您可以在println上方看到完全打印我所需的内容。但是，如何将这个字符串写入文件而不是控制台？

Answer 1

您可以使用以下代码

task writeToFile {
  // sample list.(you already have it as extra.files.collect { file -> file.name })
  List<String> sample = [ 'line1','line2','line3' ] as String[]  
  // create the folders if it does not exist.(otherwise it will throw exception)
  File extraFolder = new File( "${project.buildDir}/classes/extra")
  if( !extraFolder.exists() ) {
    extraFolder.mkdirs()
  }
  // create the file and write text to it.
  new File("${project.buildDir}/classes/extra/list.txt").text = sample.join ("\n")
}

Answer 2

一种实现方法是定义一个自定义任务，该任务将从 extra 配置生成此“索引”文件，并使现有的processResources任务依赖于此自定义任务

类似的方法会起作用：

// Task that creates the index file which lists all extra libs
task createExtraFilesIndex(){
    // destination directory for the index file
    def indexFileDir = "$buildDir/resources/main"
    // index filename
    def indexFileName = "extra-libs.index"
    doLast{
        file(indexFileDir).mkdirs()
        def extraFiles = configurations.extra.filter { file -> file.isFile () }
        // Groovy concise syntax for writing into file; maybe you want to delete this file first.
        file( "$indexFileDir/$indexFileName") << extraFiles.files.collect { file -> file.name }.join ("\n")
    }
}

// make  processResources depends on createExtraFilesIndex task
processResources.dependsOn createExtraFilesIndex

在Gradle构建脚本中以编程方式创建文件

2 个答案: