Convert from scala.collection.Seq<String> to java.util.List<String> in Java code

问题

I'm calling a Scala method, from Java. And I need to make the conversion from Seq to List.

I can't modified the signature of the Scala method, so I can't used the asJavaCollection method from scala.collection.JavaConversions._

Any ideas of how can I achieve this?

Using Scala 2.9.3

回答1:

You're on the right track using JavaConversions, but the method you need for this particular conversion is seqAsJavaList:

java.util.List<String> convert(scala.collection.Seq<String> seq) {
    return scala.collection.JavaConversions.seqAsJavaList(seq);
}

Update: JavaConversions is deprecated, but the same function can be found in JavaConverters.

java.util.List<String> convert(scala.collection.Seq<String> seq) {
    return scala.collection.JavaConverters.seqAsJavaList(seq);
}

回答2:

Since Scala 2.9, you shouldn't use implicits from JavaConversions since they are deprecated and will soon be removed. Instead, to convert Seq into java List use convert package like this (although it doesn't look very nice):

import scala.collection.convert.WrapAsJava$;

public class Test {
    java.util.List<String> convert(scala.collection.Seq<String> seq) {
        return WrapAsJava$.MODULE$.seqAsJavaList(seq);
    }
}

回答3:

Since 2.12 this is the recommended way:

public static <T> java.util.List<T> convert(scala.collection.Seq<T> seq) {
    return scala.collection.JavaConverters.seqAsJavaList(seq);
}

All other methods a @deprecated("use JavaConverters or consider ToJavaImplicits", since="2.12.0")

回答4:

(In case you want to do this conversion in Scala code)

You can use JavaConverters to make this really easy.

import collection.JavaConverters._
val s: Seq[String] = ...
val list: java.util.List<String> = s.asJava

来源：https://stackoverflow.com/questions/17737631/convert-from-scala-collection-seqstring-to-java-util-liststring-in-java-code