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Basic Gradle build file for a project using GroovyFX 0.2
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apply plugin:'groovy' | |
project.ext.set('javafxHome', System.env['JAVAFX_HOME'] | |
repositories { mavenCentral() } | |
dependencies { | |
groovy 'org.codehaus.groovy:groovy-all:1.8.6' | |
compile 'org.codehaus.groovyfx:groovyfx:0.4.0' | |
compile files("${javafxHome}/rt/lib/jfxrt.jar") | |
} | |
task run(type: JavaExec) { | |
main = 'helloGroovyFX' | |
classpath sourceSets.main.runtimeClasspath | |
} | |
task makeDirs(description:'make all dirs for project setup') << { | |
def sources = [sourceSets.main, sourceSets.test] | |
sources*.allSource*.srcDirs.flatten().each { File srcDir -> | |
println "making $srcDir" | |
srcDir.mkdirs() | |
} | |
} | |
task wrap(type:Wrapper, description:"create a gradlew") { | |
gradleVersion = '1.0-rc-3' | |
} |
if javafx lib is already on the path you don't need javafxHome
, as the version groovyfx:8.0.0 this is working for me:
apply plugin: 'groovy'
repositories {
jcenter()
}
dependencies {
compile 'org.groovyfx:groovyfx:8.0.0'
}
task run(type: JavaExec) {
main = 'helloGroovyFX'
classpath sourceSets.main.runtimeClasspath
}
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newer versions of gradle (since 1.0-milestone-9) prefer the syntax
project.ext.set('javafxHome', System.env['JAVAFX_HOME'])
instead of setting a dynamic property (line 3).