Gradle Javaプロジェクトclasspathの設定

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Running java class from gradle build script
Old Forum
u1234
tulsoba
Nov '12
I need to run a main method on a java class from my build.gradle file. But how do avoid to specify the whole classpath manually like I am doing below:

task(runSimple, dependsOn: 'classes', type: JavaExec) {
 main = 'com.MyMain'
 classpath=sourceSets.main.runtimeClasspath
 classpath+=sourceSets.test.runtimeClasspath
 classpath+=sourceSets.main.resources
 classpath+=configurations.compile.dependencies // all dependencies for this project - is empty :-(
}

The above does not work because the classpath does not contain the projects dependencies. Is there a way to specify that the classpath should be the one used when building this project which also contains all necessary dependencies?
If I add the following to my build.gradle file:

configurations.compile.dependencies.each { println it }

nothing is printed even though running this from commandline:

gradle dependencies

prints a dozen for the compile configuation.

created Nov '12

last reply Jun '13

4 replies

6.8k views
4 users
5 links

mbjarland1
Matias Bjarland
Nov '12
Haven't tested this so a bit of freestyling here, but looking at the JavaExec API doc at:
206http://gradle.org/docs/current/dsl/org.gradle.api.tasks.JavaExec.html#org.gradle.api.tasks.JavaExec:classpath206
we can see that the type for the 'classpath' property is FileCollection6. The DSL guide further tells us that the Project.configurations property is of type ConfigurationContainer3 and that expressions like:

configurations.compile

return values of type Configuration7. Reading the docs on this last link we can see that, quote:
Configuration is an instance of a FileCollection that contains all dependencies...
Ok, so what we need for the JavaExec.classpath is something of type FileCollection and that is exactly what the "configurations.compile"expression returns.
So to sum this up, I think the following should work for you:

task runSimple(type: JavaExec, dependsOn: 'classes') {
    main = 'com.MyMain'
    classpath = configurations.runtime
}

replacing the 'runtime' part with 'compile' or whichever classpath you might need in your case,
luke_daley
Nov '12
@u1234: does Matias' response answer your question?
u1234
tulsoba
Nov '12
Not really. I can only make it work if I do it in the afterEvaluate scope and on the main sourceSet:

afterEvaluate{
  main = 'com.MyMain'
  classpath=sourceSets.main.runtimeClasspath
//
classpath=configurations.runtime , does not work
  println "the classpath: "
  classpath.each { println it }
 }

With the above it finds the class but I get another error that a .properties file cannot be found which is located the test resources (I know that this should only be used when running tests):

test {
  java {
   srcDir 'test'
  }
  resources {
   srcDir 'testResources'
  }
 }

Therefore I have added the test resources like this:

afterEvaluate{
  main = 'com.MyMain'
  classpath=sourceSets.main.runtimeClasspath
  classpath=+sourceSets.test.resources
//
classpath=configurations.runtime , does not work
  println "the classpath: "
  classpath.each { println it }
 }

but it gives an error:

groovy.lang.MissingMethodException: No signature of method: org.gradle.api.internal.file.DefaultSourceDirectorySet.positive() is applicable for argument types: ()

Any suggestions on how to include resources from the test sourceSet when running the above class?
6 months later
Gail_Stewart
gstewart
Jun '13
--create a new configuration "myMain"
-- to pickup the main classpath and the test classpath mymain.extendsFrom(test,main)
-- then classpath
classpath=configurations.myMain
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ソース:https://discuss.gradle.org/t/running-java-class-from-gradle-build-script/6065