I have few files in a resource folder and I have a mapping class for the same. Now All I want is to load each file in different config class using pureconfig. Is there any way to load it via providing only resource folder name.
- src
- main
- resources
- configs
- conf1.json
- conf2.json
I want something like this
ConfigSource.resources("configs")
and it should return
List<Conf>
The current approach is something like this
def main(args: Array[String]): Unit = {
implicit def hint[A]: ProductHint[A] =
ProductHint[A](ConfigFieldMapping(CamelCase, CamelCase))
val resourceFiles = getResourceFolderFiles("configs")
val configs = new ListBuffer[SampleConfig];
resourceFiles.foreach(file =>
configs.append(
ConfigSource
.file(file)
.load[SampleConfig]
.getOrElse(null)))
println(configs.size)
}
private def getResourceFolderFiles(folder: String): Array[File] = {
val loader = Thread.currentThread.getContextClassLoader
val url = loader.getResource(folder)
val path = url.getPath
new File(path).listFiles
}
Is there any simplest way possible?
implicit def hint[A]: ProductHint[A] =
ProductHint[A](ConfigFieldMapping(CamelCase, CamelCase))
val sampleConfigList =
Try(Thread.currentThread().getContextClassLoader.getResource("configs").getPath)
.flatMap(filePath => Try(new File(filePath).listFiles().toList))
.map(fileList =>
fileList.flatMap(file => ConfigSource.file(file).load[SampleConfig].toOption)
)
.getOrElse(List.empty[SampleConfig])