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Slackbot

03/21/2022, 2:38 PM

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ephemient

03/21/2022, 2:45 PM

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// gradle/libs.versions.toml
[libraries]
group-name = "group:name:version"

// build.gradle
dependencies {
    configurationName(libs.group.name) {
        artifact {
            type = "extension"
        }
    }
}

I believe you mean "extension" rather than "classifier", because that's what "@" does

joschi

03/21/2022, 2:54 PM

Thanks a lot. I really mean classifier, for example:

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testImplementation("org.apache.kafka:kafka-clients:3.0.0:test")

joschi

03/21/2022, 3:00 PM

Unfortunately this doesn't seem to work. TOML:

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[versions]
kafka = "2.8.1"

[libraries]
kafka-clients = { module = "org.apache.kafka:kafka-clients", version.ref = "kafka" }

build.gradle

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api(libs.kafka.clients) {
      artifact {
        classifier = "test"
      }
    }

Error message with Gradle 7.4.1:

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* What went wrong:
A problem occurred evaluating project ':my-project'.
> Could not resolve all dependencies for configuration ':my-project:runtimeClasspath'.
   > Could not find method artifact() for arguments [build_3bwyqt3y3wr3wnc2ldw9wgwci$_run_closure2$_closure3$_closure6$_closure7@675386e7] on constraint org.apache.kafka:kafka-clients:2.8.1, attributes=null of type org.gradle.api.internal.artifacts.dependencies.DefaultDependencyConstraint.

joschi

03/21/2022, 3:04 PM

Ah, wait. I think I misunderstood how to use this in a Java platform (or rather not use it in a Java platform).

Thomas Broyer

03/21/2022, 3:29 PM

Note that ideally you wouldn't have to explicitly use classifiers and/or extensions from dependency declarations. Dependency selection should be based on variants, according to variant attributes, and you can use component metadata rules to declare variants on dependencies that lack them. https://docs.gradle.org/current/userguide/component_metadata_rules.html#making_variants_published_as_classified_jars_explicit https://docs.gradle.org/current/userguide/java_testing.html#publishing_test_fixtures

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