如何从 Java 列表中获取 Scala 列表？
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How to get Scala List from Java List?
提问by ace
I have a Java API that returns a List like:
我有一个 Java API，它返回一个列表，如：
public List<?> getByXPath(String xpathExpr)
I am using the below scala code:
我正在使用以下 Scala 代码：
val lst = node.getByXPath(xpath)
Now if I try scala syntax sugar like:
现在，如果我尝试像下面这样的 Scala 语法糖：
lst.foreach{ node => ... }
it does not work. I get the error:
这是行不通的。我收到错误：
value foreach is not a member of java.util.List[?0]
It seems I need to convert Java List to Scala List. How to do that in above context?
看来我需要将 Java 列表转换为 Scala 列表。在上述情况下如何做到这一点？
采纳答案by ace
EDIT: Note that this is deprecated since 2.12.0. Use JavaConvertersinstead. (comment by @Yaroslav)
编辑：请注意，自 2.12.0 以来已弃用。使用JavaConverters来代替。（@Yaroslav 的评论）
Since Scala 2.8this conversion is now built into the language using:
从Scala 2.8 开始，这种转换现在使用以下方法内置到语言中：
import scala.collection.JavaConversions._
...
lst.toList.foreach{ node => .... }
works. asScaladid not work
作品。asScala不工作
回答by montreal.eric
There's a handy Scala object just for this - scala.collection.JavaConverters
有一个方便的 Scala 对象只是为了这个 - scala.collection.JavaConverters
You can do the import and asScalaafterwards as follows:
您可以asScala按如下方式进行导入和之后的操作：
import scala.collection.JavaConverters._
val lst = node.getByXPath(xpath).asScala
lst.foreach{ node => .... }
This should give you Scala's Bufferrepresentation allowing you to accomplish foreach.
这应该为您提供 Scala 的Buffer表示，允许您完成foreach.
回答by Radix
I was looking for an answer written in Java and surprisingly couldn't find any clean solutions here. After a while I was able to figure it out so I decided to add it here in case someone else is looking for the Java implementation (I guess it also works in Scala?):
我正在寻找用 Java 编写的答案，但令人惊讶的是在这里找不到任何干净的解决方案。过了一会儿，我弄清楚了，所以我决定在这里添加它，以防其他人正在寻找 Java 实现（我猜它也适用于 Scala？）：
JavaConversions.asScalaBuffer(myJavaList).toList()
回答by Leandro carrasco
If you have to convert a Java List<ClassA>to a Scala List[ClassB], then you must do the following:
如果必须将 Java 转换为List<ClassA>Scala List[ClassB]，则必须执行以下操作：
1) Add
1) 添加
import scala.collection.JavaConverters._
2) Use methods asScala, toListand then map
2）使用方法asScala，toList然后map
List <ClassA> javaList = ...
var scalaList[ClassB] = javaList.asScala.toList.map(x => new ClassB(x))
3) Add the following to the ClassBconstructor that receives ClassAas a parameter:
3）将以下内容添加到作为参数ClassB接收的构造ClassA函数中：
case class ClassB () {
?? def this (classA: ClassA) {
????? this (new ClassB (classA.getAttr1, ..., classA.getAttrN))
?? }
}
回答by user1553728
Since scala 2.8.1 use JavaConverters._to convert scala and Java collections using asScala and asJava methods.
由于 scala 2.8.1 用于JavaConverters._使用 asScala 和 asJava 方法转换 scala 和 Java 集合。
import scala.collection.JavaConverters._
import scala.collection.JavaConverters._
javalist.asScala
javalist.asScala
scalaSeq.asJava
scalaSeq.asJava
see the Conversion relationship scala doc site
请参阅转换关系scala doc 站点
回答by Xing-Wei Lin
Shortcut to convert java list to scala list
将java列表转换为scala列表的快捷方式
import scala.collection.JavaConverters._
import scala.collection.JavaConverters._
myjavaList.asScala.toList
myjavaList.asScala.toList
