我正在尝试在 Scala 中递归解压缩文件,我已将现有的 Java 代码修改为 scala 语法。
在我的代码中,当我声明一个字节数组来读取数据时,我收到以下错误:类型不匹配;找到:Array[java.lang.Byte] 必需:Array[scala.Byte]
我的 inputstream.read 函数还给我一个错误:用替代方法读取重载方法值: (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造成的。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());
}
}
}
}
最佳答案
尝试删除该字符串
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)
关于java - Scala 字节数组类型不匹配错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43319161/