在Pentaho中解压文件问题

我试图在作业中解压一个文件，一切正常，直到zip中的文件名有一些特殊字符，如“á，é，í，ó，ú”。当zip中的文件名包含这些字符时，我得到一个错误，并记录以下日志：

Unzip file - ERROR (version 8.1.0.0-365, build 8.1.0.0-365 from 2018-04-30 09.42.24 by buildguy) : Could not unzip file [file:///C:/pentaho/data/example.zip]. Exception : [MALFORMED]
Unzip file - ERROR (version 8.1.0.0-365, build 8.1.0.0-365 from 2018-04-30 09.42.24 by buildguy) : java.lang.IllegalArgumentException: MALFORMED
Unzip file -    at java.util.zip.ZipCoder.toString(Unknown Source)
Unzip file -    at java.util.zip.ZipFile.getZipEntry(Unknown Source)
Unzip file -    at java.util.zip.ZipFile.access$900(Unknown Source)
Unzip file -    at java.util.zip.ZipFile$ZipEntryIterator.next(Unknown Source)
Unzip file -    at java.util.zip.ZipFile$ZipEntryIterator.nextElement(Unknown Source)
Unzip file -    at java.util.zip.ZipFile$ZipEntryIterator.nextElement(Unknown Source)
Unzip file -    at org.apache.commons.vfs2.provider.zip.ZipFileSystem.init(ZipFileSystem.java:83)
Unzip file -    at org.apache.commons.vfs2.provider.AbstractVfsContainer.addComponent(AbstractVfsContainer.java:49)
Unzip file -    at org.apache.commons.vfs2.provider.AbstractFileProvider.addFileSystem(AbstractFileProvider.java:96)
Unzip file -    at org.apache.commons.vfs2.provider.AbstractLayeredFileProvider.createFileSystem(AbstractLayeredFileProvider.java:80)
Unzip file -    at org.apache.commons.vfs2.provider.AbstractLayeredFileProvider.findFile(AbstractLayeredFileProvider.java:56)
Unzip file -    at org.apache.commons.vfs2.impl.DefaultFileSystemManager.resolveFile(DefaultFileSystemManager.java:711)
Unzip file -    at org.pentaho.di.core.vfs.ConcurrentFileSystemManager.resolveFile(ConcurrentFileSystemManager.java:91)
Unzip file -    at org.apache.commons.vfs2.impl.DefaultFileSystemManager.resolveFile(DefaultFileSystemManager.java:648)
Unzip file -    at org.pentaho.di.core.vfs.KettleVFS.getFileObject(KettleVFS.java:152)
Unzip file -    at org.pentaho.di.core.vfs.KettleVFS.getFileObject(KettleVFS.java:109)
Unzip file -    at org.pentaho.di.job.entries.unzip.JobEntryUnZip.unzipFile(JobEntryUnZip.java:626)
Unzip file -    at org.pentaho.di.job.entries.unzip.JobEntryUnZip.processOneFile(JobEntryUnZip.java:525)
Unzip file -    at org.pentaho.di.job.entries.unzip.JobEntryUnZip.execute(JobEntryUnZip.java:470)
Unzip file -    at org.pentaho.di.job.Job.execute(Job.java:676)
Unzip file -    at org.pentaho.di.job.Job.execute(Job.java:817)
Unzip file -    at org.pentaho.di.job.Job.execute(Job.java:493)
Unzip file -    at org.pentaho.di.job.Job.run(Job.java:380)

如何解决此问题

我正在附上这份工作的图片：

警察局。我已经看过这里和其他论坛了。

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谢谢

我找到了解决方案，可以帮助某人，所以我将其发布

步骤：获取变量，以获取所需的参数

步骤：用户定义的Java类，在这里我解压缩文件，更改编码，这是代码：

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步骤：设置变量

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您正在使用的操作系统是什么？您正在写入的文件系统是什么？如果知道创建zip文件的操作系统和文件系统是什么？我使用的是Windows 10，它应该是NTFS。让我查一下其他信息。当文件名没有这些字符时，它可以正常工作。有什么建议吗？谢谢可能的副本？

import javax.swing.*;
import java.io.File;
import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Enumeration;
import org.apache.commons.compress.archivers.zip.ZipArchiveEntry;
import org.apache.commons.compress.archivers.zip.ZipFile;
import org.apache.commons.io.IOUtils;


public boolean processRow(StepMetaInterface smi, StepDataInterface sdi) throws KettleException
{

    Object[] r = getRow();
    if (r == null) {
        setOutputDone();
        return false;
    }
    Object[] outputRow = createOutputRow(r, data.outputRowMeta.size());


    String fname = getVariable("VARIABLE_NAME", null);
    String outDir = getVariable("VARIABLE_NAME", null);


    System.out.println(fname + "  " + outDir);

    try {
            java.io.File inputFile = new java.io.File(fname);
            ZipFile zipFile = new ZipFile(inputFile, "cp866", false);
            Enumeration enumEntry = zipFile.getEntries();
            int i = 0;
            while(enumEntry.hasMoreElements()){
                ZipArchiveEntry entry = (ZipArchiveEntry) enumEntry.nextElement();
                String entryName = entry.getName();
                System.out.println(entryName);
                OutputStream os = new FileOutputStream(new File(outDir, entryName));
                InputStream is = zipFile.getInputStream(entry);
                IOUtils.copy(is, os);


                is.close();
                os.close();
                //Printing output fields
                get(Fields.Out, "FNAME").setValue(outputRow, fname);
                get(Fields.Out, "FileNameUnzipped").setValue(outputRow, entryName);
                putRow(data.outputRowMeta, outputRow);
            }
        } catch (Exception exc) {
            System.out.println("Faild to unzip");
            exc.printStackTrace();
        }


    return true;    
}