Unzip zip

Hi all,
I am trying to unzip in sender CC and Zip at Receiver CC.
I am using AF_Modules/PayLoadZipBean at both sender and receiver cc's
I am getting the error below
Attempt to process file failed with com.sap.engine.services.jndi.persistent.exceptions.NameNotFoundException: Object not found in lookup of PayLoadZipBean
what could be the reason for this error?
Thanks,
Srinivas

Hi,
Use adapter module PayloadZipBean to zip/unzip.
/people/stefan.grube/blog/2007/02/20/working-with-the-payloadzipbean-module-of-the-xi-adapter-framework
http://help.sap.com/saphelp_nw04/helpdata/en/45/da9358a1772e97e10000000a155369/frameset.htm
Thanks,
Satya Kumar..

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Easy way to unzip zipped files?

I'm trying to figure out an easy way to unzip zipped files. I'm using C# in VS 4.5.
I download this:
http://dotnetzip.codeplex.com/
I couldn't get any dlls installed. I couldn't set any reference to anything.
I also tried to follow the example here:
http://www.danderson.me/posts/zipfile-class-system-io-compression-filesystem/
I can't find anything titled 'System.IO.Compression.dll'. I set a reference to these two:
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For instance, I think this should work:
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using System.IO;
using System.IO.Compression;
namespace ConsoleApplication
class Program
static void Main(string[] args)
string startPath = @"c:\example\start";
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string extractPath = @"c:\example\extract";
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ZipFile.ExtractToDirectory(zipPath, extractPath);
However, i keep getting an error message that says 'The name 'ZipFile' does not exist in the current context'.
Knowledge is the only thing that I can give you, and still retain, and we are both better off for it.

Hi
Please include the below namespace and try it out.
System.IO.Compression.FileSystem
The code would be like this.
using System;
using System.IO;
namespace ConsoleApplication
class Program
static void Main(string[] args)
string startPath = @"c:\example\start";
string zipPath = @"c:\example\result.zip";
string extractPath = @"c:\example\extract";
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System.IO.Compression.FileSystem.dll
Please let me know in case of any questions you may have around this. Thank you.

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Message was edited by: CtDave

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Hi,
I found many ways to unzip a ZIP archive,
but which is the best way in term of performance ?
I found some examples doing :
ZipFile zipFile = new ZipFile( file ) ;
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Message was edited by:
matt.webdev
Message was edited by:
matt.webdev
Message was edited by:
matt.webdev

Hi there,
Pl see code below. For example, create a file named "interest"on your c:\test\interest in notepad with your recordkey line info.
recordKey     dollarAmount     interestRate     years
1234          50          3          4
5678          60          4          5
9876          70          5          6
8765          80          6          7
program below extracts recordKey line info if available in file.
import java.io.*;
public class SearchString
     String message = "9999";
     public static void main(String args[])
          SearchString ss = new SearchString();
          ss.searchFile("1234");
     public String searchFile(String sRecordKey)
          try
               BufferedReader br = new BufferedReader(new FileReader("c:/test/interest.txt"));
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                         // passing a specific recordKey to search
                         if (str.indexOf("5678") != -1)
                                   System.out.println(str);
                                   message = str;
          catch (IOException e)
               System.out.println("Could Not Search File");
          return message;

Unzip - zip file

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my environment is...
IBM websphere
Struts frame work.
My uploaded zip file contains some excel sheets and some images.
each image name is mentioned in excel sheet.
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My each excel sheet contains a Learning Course for a online exam with couple of Questions.
each Question in may containg some image.
so, my problem is uploading the contents of course in excel sheet to database along with images names as
course1_question1_imagename.gif
I am expecting some suggestions and help from the FORUMS.

I am expecting some suggestions and help from the
FORUMS.
good luck

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To compress items:
Do any of the following:
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Control-click an item and choose Compress from the shortcut menu.
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If you compress a single item, the compressed file has the name of the original item with the .zip extension. If you compress multiple items at once, the compressed file is called Archive.zip.
To uncompress (expand) an item:
Double-click the .zip file.
Basically you can just right click (control click) on a folder or group of files and compress them. To uncompress just a simple double click will do it.
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Solved
All you need to do is go to: ~/Library/Preferences/ (⇧⌘G)
find and delete this plist: com.apple.archiveutility.plist
Restart your mac.
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Howto use zipped-file for payload?

hi experts,
in our interface, we get a message with a zipped xml-document. so i have to change the "main payload" to the attachment and then unzip the file.
to swap the payload i used the following entry in the "modul"-tab:
AF_Modules/PayloadSwapBean, local enterprise bean, transform
keys:
transform, swap.keyName, payload-name
transform, swap.keyValue, attachment-1
--> i think this works, because in the response message i see only the payload "attachment-1"
but i am not able to unzip the file.
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AF_Modules/PayloadZipBean, local enterprise bean, unzip
keys:
unzip, zip.mode, unzip
what am i doing wrong, or did i forgot something?
thx 4 your help!

Hi dieter,
in your sender CC (tab module), try that:
Processing Sequence:
Nbr     Module Name                     Type                    Module Key
1     AF_Modules/PayloadZipBean     Local Enterprise Bean     unzip
2     AF_Modules/PayloadSwapBean     Local Enterprise Bean     swappayload
3     CallSapAdapter                     Local Enterprise Bean     0
Module Configuration:
Module Key              Parameter Name     Parameter Value
swappayload          swap.keyName     content-type
swappayload          swap.keyValue     application/xml
unzip                     zip.mod         unzip
Regards.
Mickael

Problem in unzip a file

hai,
Iam trying to unzip a file but its not working.
//Unzip a file
final int BUFFER = 2048;
String zfname=application.getRealPath("/") + "temp/Bulk.zip";
       File zf=new File(zfname);
try {
         BufferedOutputStream dest = null;
         FileInputStream fis = new FileInputStream(zf);
         ZipInputStream zis = new ZipInputStream(new BufferedInputStream(fis));
         ZipEntry entry;
         while((entry = zis.getNextEntry()) != null)
                                   System.out.println("Extracting: " +entry);
                 int count;
                 byte data[] = new byte[BUFFER];
                // write the files to the disk
                File outFile=new File(outPath);
                //FileOutputStream fos = new FileOutputStream(outFile);
                FileOutputStream fos = new FileOutputStream(entry.getName());
                dest = new BufferedOutputStream(fos, BUFFER);
                while ((count = zis.read(data, 0, BUFFER))!= -1)
                    dest.write(data, 0, count);
                dest.flush();
                dest.close();
         zis.close();
      } catch(Exception e) {
         e.printStackTrace();
Bulk.zip folder has the following files:
Bulk/test.txt
Bulk/fm.jsp
I get the following exception
java.io.FileNotFoundException: Bulk\test.txt
(The system cannot find the path specified)
Iam i making mistake in this line
FileOutputStream fos = new FileOutputStream(entry.getName());
                dest = new BufferedOutputStream(fos, BUFFER);Thanks,
Thanuja.

ok vijay iam so sorry for not mentioning how i rectified it. somehow i missed it. may be was bit excited when i got the result.
anyways below code worked.
//Unzip a file
    String zip_fname=request.getParameter("zfilename")!=null?request.getParameter("zfilename"):"";
    String fromZip=zip_fname;
    //out.println("FromZip:"+fromZip);
    File zipFname=new File(newFileName);
    String toLocation=newFilePath; //path to unzip the file
    String seperator = System.getProperty("file.separator");
    System.out.println("unzipping zip file to "+toLocation);
    try
        BufferedOutputStream dest = null;
        File zipDir = new File(toLocation);
         zipDir.mkdir();
        ZipFile zip = new ZipFile(fromZip);
        final int BUFFER = 2048;
        FileInputStream fin=new FileInputStream(fromZip);
        ZipInputStream zis = new ZipInputStream(new BufferedInputStream(fin));
        ZipEntry entry;
        while((entry = zis.getNextEntry()) != null)
                 //System.out.println("Extracting: " +entry);
                 int count;
                 byte data[] = new byte[BUFFER];
                 String zip_Fname=entry.getName().substring(entry.getName().lastIndexOf("/")+1);
                    if(!zip_Fname.equals(""))
                        zipFiles.add(zip_Fname);
              if(!zip_Fname.equals(""))
                 FileOutputStream fos = new FileOutputStream(toLocation + seperator + entry.getName().substring(entry.getName().lastIndexOf("/")+1));
                 dest = new BufferedOutputStream(fos, BUFFER);
                 while ((count = zis.read(data, 0, BUFFER))!= -1)
                    dest.write(data, 0, count);
                dest.flush();
                dest.close();
         zis.close();
    catch (Exception ex)
        unZipResult++;
        System.out.println(ex);
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Thanuja.

Unzip file

Using this code:
Inflater inflater = new Inflater();
FileInputStream fis = new FileInputStream("c:\\a\\unzip.zip"); //args[0].toString()
BufferedInputStream bis = new BufferedInputStream(fis);
ByteArrayOutputStream bos = new ByteArrayOutputStream();
int len;
while ((len = bis.read()) != -1)
bos.write(len);
byte[] input = bos.toByteArray();
byte[] output = new byte[2000000];
inflater.setInput(input);
inflater.inflate(output);
to inflate a ZIP file I got following Exception:
java.util.zip.DataFormatException: unknown compression method
at java.util.zip.Inflater.inflateBytes(Native Method)
at java.util.zip.Inflater.inflate(Inflater.java:195)
at java.util.zip.Inflater.inflate(Inflater.java:212)
at UnZIP.UnZIP.main(UnZIP.java:27)
Exception in thread "main"
What's wrong with my code???
Thanks
Frank

Take a look at class ZipFile.
o Create one on your input.
o Get an enumeration of the entries - the headers for your files within the zip file.
o Ask for the input stream of the file you want.
o Read the data - it is unzipped for you.

Unzip zip

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