scala - get class from a string

Question

From Scala, I'm using a Java library that expects a class argument. Example:

def service: OAuthService  = new ServiceBuilder()
                                    .provider(classOf[RunApi])

RunApi is Java class.

I'd like to be able pass a variety of classes to provider though. I have a list of them in String format.

Example, if I know the RunApi in String format; e.g. "com.me.RunApi", how can construct the equivalent to above code?

om-nom-nom · Accepted Answer · 2012-02-13 22:22:20Z

19

Use forName method:

scala> Class.forName("scala.collection.mutable.ArrayBuffer")
res0: java.lang.Class[_] = class scala.collection.mutable.ArrayBuffer

answered Feb 13, 2012 at 22:22

om-nom-nom

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3 Comments

Todd M Over a year ago

Thanks, thought it was going to work, but: [error] (org.scribe.builder.api.Api)org.scribe.builder.ServiceBuilder <and> [error] (java.lang.Class[_ <: org.scribe.builder.api.Api])org.scribe.builder.ServiceBuilder [error] cannot be applied to (java.lang.Class[?0(in method service)]) [error] .provider(Class.forName("com.me.RunApi")) [error] ^ [error] one error found

Alexey Romanov Over a year ago

Just cast it to the required type: Class.forName("com.me.RunApi").asInstanceOf[Class[_ <: org.scribe.builder.api.Api]].

Todd M Over a year ago

Thanks. I tried that as well as Class.forName("com.me.RunApi").asInstanceOf(Api), but: java.lang.Class cannot be cast to org.scribe.builder.api.Api. RunApi extends an abstract parent class which implements Api interface.

Bday · Accepted Answer · 2022-01-20 13:45:36Z

2

val yourInstance: ClassOrTrait = Class.forName("Package.ClassName").newInstance().asInstanceOf[ClassOrTrait]

answered Jan 20, 2022 at 13:45

Bday

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