Wud*_*ong 9 java groovy gradle
我想使用gradle的CreateStartScripts Task来生成启动应用程序的脚本.
我用以下方式使用它:
apply plugin: 'java'
mainClass = 'UIMain';
dependencies {
compile fileTree(dir: 'libs', include: '*.jar')
}
task copyResources(type: Copy) {
from 'config.ini'
into 'build/dist'
}
task copyLibs(type: Copy) {
from configurations.default
from configurations.default.allArtifacts.files
into 'build/dist/libs'
}
task generateScript(type: CreateStartScripts) {
applicationName = "Dagang"
mainClassName = mainClass
outputDir = "build/dist/"
classpath = ""
}
task distribute(dependsOn: [
build,
copyLibs,
copyResources,
generateScript
]) <<{
description = 'Copies all the project libs to the distribution place.'
}
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但是,当我运行构建时,它会遇到如下错误:
评估根项目'dagang'时出现问题.[org.gradle.BuildExceptionReporter]原因:在根项目'dagang'上找不到属性'CreateStartScripts'.
任何人都可以给我一些光明吗?谢谢.
要么导入类 ( org.gradle.api.tasks.application.CreateStartScripts
),要么使用application
插件。后者通常是优选的。
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