How to convert properties file to JSON using scala
Properties file contains
a.b.10=C
a.b.11=C50
a.b.12=C508
Output should be {"a":{"b":{"10":"C","11":"C50","12":"C508"}}}
You can use circe-config
. Example:
import io.circe.config.parser.parse
val result = parse("""
a.b.10=C
a.b.11=C50
a.b.12=C508
""").map(_.noSpaces)
The example above will produce an Either[ParsingFailure, String]
, which you can then destructure to handle failures, for example
result match {
case Left(failure) => // handle parsing failure
case Right(jsonString) => // do something with your json string
}
The json string produced by the example above is:
{"a":{"b":{"12":"C508","10":"C","11":"C50"}}}
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