如何将Java流转换为Scala流? [英] How to convert a Java Stream to a Scala Stream?
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问题描述
作为将Java代码转换为Scala代码的一部分,我需要将Java流Files.walk(Paths.get(ROOT))
转换为Scala.我无法通过谷歌搜索找到解决方案. asScala
不会这样做.有提示吗?
As a part of an effort of converting Java code to Scala code, I need to convert the Java stream Files.walk(Paths.get(ROOT))
to Scala. I can't find a solution by googling. asScala
won't do it. Any hints?
以下是相关代码:
import static org.springframework.hateoas.mvc.ControllerLinkBuilder.linkTo;
import static org.springframework.hateoas.mvc.ControllerLinkBuilder.methodOn;
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.stream.Collectors;
....
Files.walk(Paths.get(ROOT))
.filter(path -> !path.equals(Paths.get(ROOT)))
.map(path -> Paths.get(ROOT).relativize(path))
.map(path -> linkTo(methodOn(FileUploadController.class).getFile(path.toString())).withRel(path.toString()))
.collect(Collectors.toList()))
其中Files.walk(Paths.get(ROOT))
返回类型在Java中为Stream<Path>
.
where the Files.walk(Paths.get(ROOT))
return type is Stream<Path>
in Java.
推荐答案
还有一种更好的方法,不需要提到compat层或实验性2.11功能 @marcospereira
There is a slightly nicer way without needing the compat layer or experimental 2.11 features mentioned here by @marcospereira
基本上只使用迭代器:
import java.nio.file.{Files, Paths}
import scala.collection.JavaConverters._
Files.list(Paths.get(".")).iterator().asScala
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