如何读取属性文件并在项目Gradle脚本中使用值？

Question

我正在使用Gradle脚本工作，需要在其中读取local.properties文件并使用中的属性文件中的值build.gradle。我以下面的方式来做。我运行了以下脚本，它现在引发错误，但是它也没有执行创建，删除和复制文件的操作。我试图打印变量的值，它显示正确的值。

有人可以让我知道这是否是正确的方法吗？我认为另一种方法是在中定义所有内容并在中gradle.properties使用它build.gradle。有人可以让我知道如何build.gradle从中访问属性build.properties吗？

build.gradle 文件：

apply plugin: 'java'

//-- set the group for publishing
group = 'com.true.test'

/**
 * Initializing GAVC settings
 */
def buildProperties = new Properties()
file("version.properties").withInputStream {
        stream -> buildProperties.load(stream)
}
//if jenkins build, add the jenkins build version to the version. Else add snapshot version to the version.
def env = System.getenv()
if (env["BUILD_NUMBER"]) buildProperties.test+= ".${env["BUILD_NUMBER"]}"
version = buildProperties.test
println "${version}"

//name is set in the settings.gradle file
group = "com.true.test"
version = buildProperties.test
println "Building ${project.group}:${project.name}:${project.version}"

Properties properties = new Properties()
properties.load(project.file('build.properties').newDataInputStream())
def folderDir = properties.getProperty('build.dir')
def configDir = properties.getProperty('config.dir')
def baseDir  = properties.getProperty('base.dir')
def logDir  = properties.getProperty('log.dir')
def deployDir  = properties.getProperty('deploy.dir')
def testsDir  = properties.getProperty('tests.dir')
def packageDir  = properties.getProperty('package.dir')
def wrapperDir  = properties.getProperty('wrapper.dir')


sourceCompatibility = 1.7
compileJava.options.encoding = 'UTF-8'

repositories {
     maven { url "http://arti.oven.c:9000/release" }
  }

task swipe(type: Delete) {
         println "Delete $projectDir/${folderDir}"
         delete "$projectDir/$folderDir"
         delete "$projectDir/$logDir"
         delete "$projectDir/$deployDir"
         delete "$projectDir/$packageDir"
         delete "$projectDir/$testsDir"
         mkdir("$projectDir/${folderDir}")
         mkdir("projectDir/${logDir}")
         mkdir("projectDir/${deployDir}")
         mkdir("projectDir/${packageDir}")
         mkdir("projectDir/${testsDir}")
}
task prepConfigs(type: Copy, overwrite:true, dependsOn: swipe) {
    println "The name of ${projectDir}/${folderDir} and ${projectDir}/${configDir}"
    from('${projectDir}/${folderDir}')
    into('${projectDir}/$configDir}')
    include('*.xml')
}

build.properties 文件：

# -----------------------------------------------------------------
# General Settings
# -----------------------------------------------------------------
application.name  = Admin
project.name = Hello Cool

# -----------------------------------------------------------------
# ant build directories
# -----------------------------------------------------------------
sandbox.dir = ${projectDir}/../..
reno.root.dir=${sandbox.dir}/Reno
ant.dir = ${projectDir}/ant
build.dir = ${ant.dir}/build
log.dir  = ${ant.dir}/logs
config.dir = ${ant.dir}/configs
deploy.dir  = ${ant.dir}/deploy
static.dir =  ${ant.dir}/static
package.dir = ${ant.dir}/package
tests.dir = ${ant.dir}/tests
tests.logs.dir = ${tests.dir}/logs
external.dir = ${sandbox.dir}/FlexCommon/External
external.lib.dir = ${external.dir}/libs

Answer 1

如果使用默认gradle.properties文件，则可以直接从build.gradle文件内部访问属性：

gradle.properties：

applicationName=Admin
projectName=Hello Cool

build.gradle：

task printProps {
    doFirst {
        println applicationName
        println projectName
    }
}

如果您需要访问自定义文件或访问其中包含的属性.（看起来需要这样做），则可以在build.gradle文件中执行以下操作：

def props = new Properties()
file("build.properties").withInputStream { props.load(it) }

task printProps {
    doFirst {
        println props.getProperty("application.name")
        println props.getProperty("project.name")
    }
}

请参阅Gradle文档的此部分以获取更多信息。

编辑

如果您想动态设置其中一些属性（如下面的注释中所述），则可以创建一个properties.gradle文件（名称并不重要）并在build.gradle脚本中需要它。

properties.gradle：

ext {
    subPath = "some/sub/directory"
    fullPath = "$projectDir/$subPath"
}

build.gradle

apply from: 'properties.gradle'

// prints the full expanded path
println fullPath