22

我是 Scala 的新手，我想自己写一些源代码让我变得更好。我编写了一个简单的对象（带有一个主条目）以模拟对当前目录的所有文件的“grep”调用。（我从 Eclipse Indigo 和 Debian Squeeze 启动程序）：

package com.gmail.bernabe.laurent.scala.tests

import java.io.File

import scala.io.Source

object DealWithFiles {

  def main(args:Array[String]){
    for (result <- grepFilesHere(".*aur.*"))
      println(result)
  }

  private def grepFilesHere(pattern:String):Array[String] = {
    val filesHere = new File(".").listFiles

    def linesOfFile(file:File) =
      Source.fromFile(file).getLines.toList

    for (file <- filesHere;
        if file.isFile
    )
      yield linesOfFile(file)(0)
  }

}

但我得到一个 java.nio.charset.MalformedInputException，我无法解决：

Exception in thread "main" java.nio.charset.MalformedInputException: Input length = 1
at java.nio.charset.CoderResult.throwException(CoderResult.java:260)
at sun.nio.cs.StreamDecoder.implRead(StreamDecoder.java:319)
at sun.nio.cs.StreamDecoder.read(StreamDecoder.java:158)
at java.io.InputStreamReader.read(InputStreamReader.java:167)
at java.io.BufferedReader.fill(BufferedReader.java:136)
at java.io.BufferedReader.readLine(BufferedReader.java:299)
at java.io.BufferedReader.readLine(BufferedReader.java:362)
at scala.io.BufferedSource$BufferedLineIterator.hasNext(BufferedSource.scala:67)
at scala.collection.Iterator$class.foreach(Iterator.scala:772)
at scala.io.BufferedSource$BufferedLineIterator.foreach(BufferedSource.scala:43)
at scala.collection.generic.Growable$class.$plus$plus$eq(Growable.scala:48)
at scala.collection.mutable.ListBuffer.$plus$plus$eq(ListBuffer.scala:130)
at scala.collection.TraversableOnce$class.toList(TraversableOnce.scala:242)
at scala.io.BufferedSource$BufferedLineIterator.toList(BufferedSource.scala:43)
at com.gmail.bernabe.laurent.scala.tests.DealWithFiles$.linesOfFile$1(DealWithFiles.scala:18)
at com.gmail.bernabe.laurent.scala.tests.DealWithFiles$$anonfun$grepFilesHere$2.apply(DealWithFiles.scala:23)
at com.gmail.bernabe.laurent.scala.tests.DealWithFiles$$anonfun$grepFilesHere$2.apply(DealWithFiles.scala:20)
at scala.collection.TraversableLike$WithFilter$$anonfun$map$2.apply(TraversableLike.scala:697)
at scala.collection.IndexedSeqOptimized$class.foreach(IndexedSeqOptimized.scala:34)
at scala.collection.mutable.ArrayOps.foreach(ArrayOps.scala:38)
at scala.collection.TraversableLike$WithFilter.map(TraversableLike.scala:696)   
at com.gmail.bernabe.laurent.scala.tests.DealWithFiles$.grepFilesHere(DealWithFiles.scala:20)
at com.gmail.bernabe.laurent.scala.tests.DealWithFiles$.main(DealWithFiles.scala:10)
at com.gmail.bernabe.laurent.scala.tests.DealWithFiles.main(DealWithFiles.scala)

在此先感谢您的帮助:)

4

2 回答 2

30

来自 JavaDoc：

当输入字节序列对于给定字符集不合法或输入字符序列不是合法的 16 位 Unicode 序列时抛出MalformedInputException 。

将当前编码作为参数传递给Source.fromFile方法。

于 2012-06-01T08:34:56.410 回答
3

您可以通过在代码中添加以下代码段来处理此字符编码异常

import scala.io.Codec
import java.nio.charset.CodingErrorAction


implicit val codec = Codec("UTF-8")
codec.onMalformedInput(CodingErrorAction.REPLACE)
codec.onUnmappableCharacter(CodingErrorAction.REPLACE)
于 2018-10-15T07:54:11.710 回答