在Pentaho-spoon中解压缩文件问题

我找到了解决方案，可以帮助某人，所以我发布了它。

1. 
   步骤：获取变量，获取我需要的参数。
   


2. 
   步骤：用户定义的Java类，在这里我解压缩文件，改变编码，这是代码：
   

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---




    
      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;    
}
</code>
  





1. 
   步骤：设置变量。
   

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