Access Play Configuration in Plain Scala Module

Question

I have a Play app that I want to split into:

rest - Play module that holds the rest interface
services - plain Scala module that holds actors and external services clients (ex S3)

I want to create a S3Client object class, but when instantiating it I need to read the config values from application.conf. I can not use play.api.Play.current.configuration since this is not a play module.

Am I thinking this wrong? What would be the most elegant way to achieve what I want?

Maybe a better approach would be to factor out all configuration stuff common to all modules to seperate module — maks, Mar 17 '14 at 13:13

score 1 · Accepted Answer · answered Mar 17 '14 at 14:26

1

just use typesafe Config factory

   import com.typesafe.config.ConfigFactory

   val conf = ConfigFactory.load("/path/to/your/conf/file/application.conf")

   //init your object with the conf file.

answered Mar 17 '14 at 14:26

Nimrod007

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score 0 · Answer 2 · edited Jan 14 '17 at 01:36

0

You don't even need to put a path in it.

If the application.conf is in the root directory, the config factory will automatically check if it's there.

edited Jan 14 '17 at 01:36

zx485

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answered Jan 14 '17 at 01:12

Access Play Configuration in Plain Scala Module

2 Answers2