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fix: Adjust the Sentry release property format according to the docs #193

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Mar 26, 2020
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fix: Adjust the Sentry release property format according to the docs #193

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10 changes: 7 additions & 3 deletions sentry/src/main/scala/com/avast/sst/sentry/SentryModule.scala
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Original file line number Diff line number Diff line change
Expand Up @@ -26,16 +26,20 @@ object SentryModule {
}

/** Makes [[io.sentry.SentryClient]] initialized with the given config and overrides the `release` property
* with `Implementation-Version` from the `MANIFEST.MF` file inside the same JAR (package) as the `Main` class.
* with `Implementation-Title`@`Implementation-Version` from the `MANIFEST.MF` file inside the same JAR (package) as the `Main` class.
*
* This format is recommended by Sentry ([[https://docs.sentry.io/workflow/releases]])
* because releases are global and must be differentiated.
*/
def makeWithReleaseFromPackage[F[_]: Sync, Main: ClassTag](config: SentryConfig): Resource[F, SentryClient] = {
for {
customizedConfig <- Resource.liftF {
Sync[F].delay {
for {
pkg <- Option(implicitly[ClassTag[Main]].runtimeClass.getPackage)
v <- Option(pkg.getImplementationVersion)
} yield config.copy(release = Some(v))
title <- Option(pkg.getImplementationTitle)
version <- Option(pkg.getImplementationVersion)
} yield config.copy(release = Some(s"$title@$version"))
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If I understand this correctly, release will not be set if ImplementationTitle or ImplementationVersion is missing, right?

I have never set Implementation-Title - maybe it's my fault. I would consider to fallback to pkg.getName.

}
}
sentryClient <- make[F](customizedConfig.getOrElse(config))
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