对于java中的集合元素并不能在scala中拿来就用的，需要进行相应的转换。

1. 转换规则如下

从下面可以看出，有些可以相互转换的，有些只能单向转换：

scala.collection.Iterable <=> java.lang.Iterablescala.collection.Iterable <=> java.util.Collectionscala.collection.Iterator <=> java.util.{ Iterator, Enumeration }scala.collection.mutable.Buffer <=> java.util.Listscala.collection.mutable.Set <=> java.util.Setscala.collection.mutable.Map <=> java.util.{ Map, Dictionary }scala.collection.mutable.ConcurrentMap <=> java.util.concurrent.ConcurrentMapscala.collection.Seq  => java.util.Listscala.collection.mutable.Seq => java.util.Listscala.collection.Set  => java.util.Setscala.collection.Map => java.util.Mapjava.util.Properties => scala.collection.mutable.Map[String, String]

 

2. 转换操作步骤

1) 导入包 import scala.collection.JavaConverters._

2) 添加 .asScala或 .asJava

 

 3. 测试案例

以下为java.util.List 和scala 中 Seq的相互转换:

def main(args: Array[String]): Unit = {        val list:java.util.List[Int] = scala2java        println("")        java2scala(list)    }    //测试Java中List转Scala中Seq    def scala2java:java.util.List[Int] = {        import scala.collection.JavaConverters._        println("scala2java")        val list = Seq(1,2,3,4).asJava        //循环输出显示        var x = 0        for(x <- Range(0,list.size)){            print(list.get(x) + ",")        }        list    }    //测试Java中List转Scala中Seq    def java2scala(list:java.util.List[Int]) = {        import scala.collection.JavaConverters._        println("java2scala")        val buffer:scala.collection.mutable.Buffer[Int] = list.asScala        //循环输出显示        buffer.foreach(x=> print(x + ","))    }

 显示输出结果：