我想使用 Gradle“应用程序”插件为第二个 mainClass 创建 startScripts。这可能吗?即使应用程序插件没有内置此功能,是否可以利用 startScripts 任务为不同的 mainClass 创建第二对脚本?
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3 回答
14
将类似这样的内容添加到您的根 build.gradle:
// Creates scripts for entry points
// Subproject must apply application plugin to be able to call this method.
def createScript(project, mainClass, name) {
project.tasks.create(name: name, type: CreateStartScripts) {
outputDir = new File(project.buildDir, 'scripts')
mainClassName = mainClass
applicationName = name
classpath = project.tasks[JavaPlugin.JAR_TASK_NAME].outputs.files + project.configurations.runtimeClasspath
}
project.tasks[name].dependsOn(project.jar)
project.applicationDistribution.with {
into("bin") {
from(project.tasks[name])
fileMode = 0755
}
}
}
然后从根或子项目按如下方式调用它:
// The next two lines disable the tasks for the primary main which by default
// generates a script with a name matching the project name.
// You can leave them enabled but if so you'll need to define mainClassName
// And you'll be creating your application scripts two different ways which
// could lead to confusion
startScripts.enabled = false
run.enabled = false
// Call this for each Main class you want to expose with an app script
createScript(project, 'com.foo.MyDriver', 'driver')
于 2014-05-19T15:57:23.233 回答
4
您可以创建多个类型的任务,CreateStartScripts
并在每个任务中配置不同的mainClassName
. 为方便起见,您可以循环执行此操作。
于 2013-08-25T22:04:01.553 回答
4
我结合了这两个答案的一部分,得出了相对简单的解决方案:
task otherStartScripts(type: CreateStartScripts) {
description "Creates OS specific scripts to call the 'other' entry point"
classpath = startScripts.classpath
outputDir = startScripts.outputDir
mainClassName = 'some.package.app.Other'
applicationName = 'other'
}
distZip {
baseName = archivesBaseName
classifier = 'app'
//include our extra start script
//this is a bit weird, I'm open to suggestions on how to do this better
into("${baseName}-${version}-${classifier}/bin") {
from otherStartScripts
fileMode = 0755
}
}
startScripts 是在应用应用程序插件时创建的。
于 2014-07-10T21:52:35.377 回答