Question

我试图在Scala中递归地解压缩文件我已经将现有的Java代码修改为scala语法。

在我的代码中，我声明一个字节数组来读取数据，我得到以下错误：类型不匹配; found：Array [java.lang.Byte] required：Array [scala.Byte]

此外，我的inputstream.read函数给出了一个错误：重载的方法值读取替换：（x $ 1：Array [scala.Byte]，x $ 2：Int，x $ 3：Int）Int（）Int （x $ 1：Array [scala.Byte]）Int不能应用于（Array [java.lang.Byte]，Int，Int）

我认为这也是由于该数组的声明。我该如何解决这个问题？有没有办法将java.lang.Byte转换为scala.Byte？

这是我的代码：

import java.io._;
import org.apache.log4j._
import org.apache.spark.SparkContext
import java.io.IOException
import scala.collection.JavaConversions._
import java.io.FileInputStream
import java.io.FileOutputStream
import java.util.zip.ZipEntry
import java.util.zip.ZipInputStream
import java.util.zip.ZipEntry
import java.util.zip.ZipFile
import java.io.InputStream
import java.io.OutputStream
import java.io.File
import java.lang.Byte

object MultiLevelUnzip 
{
    val BUFFER = 2048    
    def main (args:Array[String])
  {
    Logger.getLogger("org").setLevel(Level.ERROR)

    val sc = new SparkContext("local[*]","Unzip")
    //val Files = sc.listFiles()
    sc.stop()
  }

    def findFiles(d : File): Array[File] =
      {
        val (dirs, files) =  d.listFiles.partition(_.isDirectory)
        files ++ dirs.flatMap(findFiles)
      }

   def extractFolder(zipFile:String)= 
{
    System.out.println(zipFile);

    val file = new File(zipFile);

    val zip = new ZipFile(file);
    val newPath = zipFile.substring(0, zipFile.length() - 4);

    new File(newPath).mkdir();
    var zipFileEntries = zip.entries()

    // Process each entry
    while (zipFileEntries.hasMoreElements())
    {
        // grab a zip file entry
        val entry = zipFileEntries.nextElement()
        val currentEntry = entry.getName()
        val destFile = new File(newPath, currentEntry);
        //destFile = new File(newPath, destFile.getName());
        val destinationParent = destFile.getParentFile();

        // create the parent directory structure if needed
        destinationParent.mkdirs();

        if (!entry.isDirectory())
        {
            val is = new BufferedInputStream(zip.getInputStream(entry))
            var currentByte = null
            // establish buffer for writing file


         // val buffer = Array.fill[Byte](BUFFER)(_)


            // write the current file to disk
            val fos = new FileOutputStream(destFile)
            val dest = new BufferedOutputStream(fos,BUFFER)


            val data = new Array[Byte](BUFFER)

            while ((currentByte = is.read(data,0, BUFFER)) != -1) {
                    dest.write(data, 0, currentByte);
                }

            dest.flush();
            dest.close();
            is.close();
        }

        if (currentEntry.endsWith(".zip"))
        {
            // found a zip file, try to open
            extractFolder(destFile.getAbsolutePath());
        }
    }
}
}

Answer 1

尝试删除此字符串

import java.lang.Byte

允许编译器在数组定义中使用scala.Byte类型

val data = new Array[Byte](BUFFER)

Scala字节数组类型不匹配错误

1 个答案: