'Exclude transitive dependency of Gradle plugin

Excluding a transitive dependency in Gradle is pretty straightforward:

compile('com.example.m:m:1.0') {
     exclude group: 'org.unwanted', module: 'x'
  }

How would we go around he situation in which we use a plugin:

apply: "somePlugin"

And when getting the dependencies we realize that the plugin is bringing some transitive dependencies of its own?

javagradle


Solution 1:[1]

You can manipulate the classpath of the buildscript itself through:

buildscript {
    configurations {
        classpath {
            exclude group: 'org', module: 'foo' // For a global exclude
        }
    }
    dependencies {
        classpath('org:bar:1.0') {
            exclude group: 'org', module: 'baz' // For excluding baz from bar but not if brought elsewhere
        }
    }
}

Solution 2:[2]

You can remove dependencies after the plugin is applied, (from a single configuration, or to all configurations) using eg. compile.exclude. Note that compile resolves to a "Configuration"; see the javadocs at Configuration.exclude .

edit

Be aware that excluding dependecies could fail, if the configuration has already been resolved.

Sample script

apply plugin: 'java-library'

repositories {
    jcenter()
}

dependencies {
    compile 'junit:junit:4.12'
    compile 'ant:ant:1.6'
    compile 'org.apache.commons:commons-lang3:3.8'
}

// remove dependencies
configurations.all {
  exclude group:'junit', module:'junit'
}
configurations.compile {
  exclude group:'org.apache.commons', module: 'commons-lang3'
}

println 'compile deps:\n' + configurations.compile.asPath

Solution 3:[3]

Here is another way to enforce your project to strictly use a specific version for the build.gradle.kts

val grpcVersion = "1.45.1"

implementation("io.grpc:grpc-stub") {
 version {
     strictly(grpcVersion)
   }
}

More info can be found at the gradle documentation: https://docs.gradle.org/current/userguide/dependency_downgrade_and_exclude.html

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Louis Jacomet
Solution 2
Solution 3 bluecheese

Related Questions