Read values from config in Scala

Question

In Scala, if I have the following config:

id = 777
username = stephan
password = DG#%T@RH

The idea is to open a file, transform it into a s

Answer 1

Best way would be to use a .conf file and the ConfigFactory instead of having to do all the file parsing by yourself:

import java.io.File
import com.typesafe.config.{ Config, ConfigFactory }

// this can be set into the JVM environment variables, you can easily find it on google
val configPath = System.getProperty("config.path")

val config = ConfigFactory.parseFile(new File(configPath + "myFile.conf"))

config.getString("username")

I usually use scalaz Validation for the parseFile operation in case the file it's not there, but you can simply use a try/catch if you don't know how to use that.