Auto unzip "zipped" files inside OWB. possible?

Scenario:
Client will be placing large fixed-delimited files in a directory w/c are in ZIPPED format. This file directory will be my source location.
Question?
Is there a way to automate the process of unzipping the files in OWB and placing it in the same directory. I am thinking of using a process flow but I cannot find any utility activity for this. Any suggestions?
Thanks
-carlo

I haven't done it for a while, but I manage to archive files from the workflow via a script.
In the workflow you have to use the ExternalProcess command.

Similar Messages

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:
System.IO.Compression.FileSystem
System.IO.Compression
That did nothing at all.
For instance, I think this should work:
using System;
using System.IO;
using System.IO.Compression;
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";
ZipFile.CreateFromDirectory(startPath, zipPath);
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";
System.IO.Compression.ZipFile.CreateFromDirectory(startPath, zipPath);
System.IO.Compression.ZipFile.ExtractToDirectory(zipPath, extractPath);
Please note that you need to add a DLL reference to the framework assembly
System.IO.Compression.FileSystem.dll
Please let me know in case of any questions you may have around this. Thank you.

Creation of zip file inside another one.

Hi,
I have a problem with the zip file creation.
I have created the zip file but now i can' t
create another zip file inside the first one.
can you suggest me how can i do it ?
thanks Laura.

Hi,
try this Fm..
ISM_GENERATE_ZIP_FILE_NAME
Arunima

How to check & unzip zip file using java

Dear friends
How to check & unzip zip file using java, I have some files which are pkzip or some other zip I want to find out the type of ZIp & then I want to unzip these files, pls guide me
thanks

How to check & unzip zip file using java, I have
ve some files which are pkzip or some other zip I
want to find out the type of ZIp & then I want to
unzip these files, pls guide meWhat do you mean "other zip"? Either they're zip archives or not, there are no different types.

When I unzip .zip files I get an alias ?

I've unzipped .zip files for years, but with Lion, unzipping a .zip file (created in Snow Leopard) does not result in a set of files, it results in a set of aliases which do not point to any file when I select "Show Original."
Any advice appreciated.

The current version of the built-in Archive Utility has problems with some zip files. Use "The Unarchiver" instead -- free in the App Store.

I'm running Firefox 7.0.1 and since the September 30 security update I can no longer unzip zipped files unless I reboot my computer first or run it in Safe Mode. Can anyone help me roll Firefox back to before the update?

I regularly download zipped files from Survey Monkey. Since the latest security update, the zipped files download to my computer, but when I try and unzip them I get a message telling me the file is in use by another application. If I reboot, I can unzip the file, or if I'm in Safe Mode when I download the file it unzips fine.
I have tried three different unzip apps and all have the same trouble. Everything was working well after the September 5 update.
I'm running Windows XP with McAfee antivirus, Spybot, Ad-Aware and Malbytes Anti-malware. All are up to date.

The URL below is to Adobe's Acrobat update page.
There you'll find each of the incremental updates to Acrobat 10.
They must be installed one at a time in sequence.
http://www.adobe.com/support/downloads/product.jsp?product=1&platform=Windows
Now, if after attempted updates fail then you'll need to speak with your IT Department.
You need your install of Acrobat 10 updated to dot version 10.1.7.
Be well...
Message was edited by: CtDave

Auto unzip archive files

Okay, after I updated my laptop to the latest version of the OS, i.e., 10.9.2, it somehow stopped auto unzipping thr archive files. How do I re-enable it? I use Chrome.
I did check Preference's in Safari and 'Open "safe" files after downloading' is still checked!
Any ideas?
Thanks!

I haven't done it for a while, but I manage to archive files from the workflow via a script.
In the workflow you have to use the ExternalProcess command.

Unzip - zip file

Can any one give some info regarding extracting a uploaded zip file on server from my web application..
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.
I need to migrate the contents in excel to database.
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

Unzipping ZIP file - Best way

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 ) ;
Enumeration entries = zipFile.entries() ;
while( entries.hasMoreElements() ) {
ZipEntry entry = ( ZipEntry )entries.nextElement() ;
// extract entry
Others :
               ZipEntry zipentry;
               zipinputstream =
new ZipInputStream(
new FileInputStream(zipFile));
               zipentry = zipinputstream.getNextEntry();
               while (zipentry != null) {
// extract entry
zipentry = zipinputstream.getNextEntry();
For extracting the entries I saw examples using BufferedInputStream and BufferedOutputStream
                         BufferedInputStream is =
                              new BufferedInputStream(zipFile.getInputStream(entry));
                         int currentByte;
                         // establish buffer for writing file
                         byte data[] = new byte[1024];
                         // write the current file to disk
                         FileOutputStream fos = new FileOutputStream(destFile);
                         BufferedOutputStream dest =
                              new BufferedOutputStream(fos, 1024);
                         // read and write until last byte is encountered
                         while ((currentByte = is.read(data, 0, 1024)) != -1) {
                              dest.write(data, 0, currentByte);
                         dest.flush();
                         dest.close();
                         is.close();
others using ZipInputStream and FileOutputStream :
zipinputstream =
new ZipInputStream(
new FileInputStream(zipFile));
                    int n;
                    FileOutputStream fileoutputstream;
                    File newFile = new File(entryName);
                    String directory = newFile.getParent();
                    if (directory == null) {
                         if (newFile.isDirectory())
                              break;
                    fileoutputstream =
                         new FileOutputStream(destinationname + entryName);
                    while ((n = zipinputstream.read(buf, 0, 1024)) > -1)
                         fileoutputstream.write(buf, 0, n);
                    fileoutputstream.close();
                    zipinputstream.closeEntry();
Can u tell me which solution to use for looping through the files (by this Enumertion or by checking nextEntry is null ?)
And for the extracting (use of BufferedInp. BufferedOutp. or ZipInputStream and FileOutputStream?) in term of performance
Thank u for any tips
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"));
               String str = "";
               while((str = br.readLine()) != null)
                         // 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;

ESA scanning for a file inside of a zip inside of another zip.

There is a new file-encrypting ransomware called CTB-Locker going around that places a malicious .cab file inside of a .zip file inside of another .zip file. Two questions:
1) How far will an ESA scan depth wise into zips inside of zips.
2) Is a .cab file considered an executable by ESA or something else? If it is considered an executable, would a block based off of this match (attachment-filetype == “Executable”) catch this type threat?
Reference to the CTB-Locker: https://www.f-secure.com/weblog/archives/00002788.html
Thanks

As an additional safe-gap measure, i would also suggest (unless of course you're receiving a lot of emails with .cab attachments normally) to put in an extra filter to quarantine emails with .cab attachments as most of the attacks seen are coming as .cab formats at the moment.
In terms of how deep it'll look in.
for .zip .rar filetypes, it will look as deep as you've set it.
So if it's a viral file inside a zip (where the viral file is an executable) it will be unable to unpack up the zip and capture the executable at the content/filter levels.
However normally the AV engine itself (mcafee and sophos) should be able to sort it out for you assuming the viral definition is available already.

How I unzip pdf files?

What version of Adobe Do i need to unzip a PDF file?

A PDF file is not a ZIP file. A ZIP file is an archive that may contain one or more files of any type.
If you need to extract a PDF file from a ZIP file archive, you can use either a third party application (such as WinZip) or simply double click on the .ZIP file inside Windows Explorer to access same.
- Dov

Read Zip File and output it to the Stream

Hi,
I really need help with this topic. I need to write a function which as input read the zip file (inside: jpeg, mp3, xml)
and as output provide the output Stream of this zip file.
public OutputStream readZipToStream(String sourceZipFile) { ....}
I've tried something....
public static OutputStream    readZipToStream(String src, HttpServletResponse response) throws Exception {
        File fsrc = null;
        ZipOutputStream out = null;
        byte buffer [] = null;
        FileInputStream in = null;
        try {
            buffer = new byte[BUFFER_CREATE];
            fsrc = new File(src);
            out = new ZipOutputStream(response.getOutputStream());
            ZipEntry zipAdd = new ZipEntry(fsrc.getName());
            zipAdd.setTime(fsrc.lastModified());
            out.putNextEntry(zipAdd);
            // Read input & write to output
            in = new FileInputStream(fsrc);
            while (true) {
                int nRead = in.read(buffer, 0, buffer.length);
                if (nRead <= 0)
                    break;
                out.write(buffer, 0, nRead);
            out.flush();
            in.close();
        } catch (IOException ioe) {
            logger.error("Zip Exception: " + ioe.getMessage());
            ioe.printStackTrace();
            throw ioe;
        return out;
    } But the problem with this code when it returns ( I called from servlet: bellow)
it ask user to save the file as servlet.jsp file. So I would have to rename it to zip file later.
What I want to achive is it would ask to save zip file from the stream as name of the original zip file (if that is possible)
<%@ page language="java" contentType="text/html; charset=ISO-8859-1"
    pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<%
OutputStream stream = null;
response.setContentType("application/zip");
target = mount_media + File.separator + "test_swf.zip";
try {
stream = ZipUtil.readZipToStream(target, response);
} catch (Exception ex) {
        System.out.println(ex.getMessage());
    } finally {
if(stream != null) {
            stream.close();
%>
<%@page import="java.io.File" %>
<%@page import="java.io.OutputStream" %>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
</body>
</html>

Hi oleg_s ,
There's a contradiction of content in your JSP :
response.setContentType("application/zip");
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">The html part of your JSP seems useless for what you mean to do. Therefore, instead of a JSP, you should use a simple servlet to send your zip file.

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);
    }Thanks,
Thanuja.

How do I create and send a zip file? I have Lion and I'm working in Word for mac

To show you how little I know about any of this, I don't even know if this is the correct forum in which to ask this question. I'm a copywriter. I have a client in France. I have 15 word documents, 15 of the same documents in PDFs and a small Excel document to send him. Simple, mutlipage Word docs. Short two-page Excell sheet. I want to create a zip file for them to send to him. Can I do it? How? I have unzipped zip files other people have sent to me, but I've never created or sent one myself. I'm working in Word for Mac 2011. I have Lion. I'm using Mac's email. Please don't assume I know anything. I need very, very simple, play by play instructions. You are all very helpful, so I have high hopes.
Nancy

Visit:
http://download.cnet.com/MacZip/3000-2250_4-10025248.html
You will be able to both zip and unzip. If the recipient of your zipped file does not have zip, he/she will need to get it, or you could send him/her the above address so that he/she can unzip his/her file.
(I wish the English language could settle on the use of something less clumsy that he/she him/her etc.)
BTW, I think that you can zip only folders, not separate files, so put even one file into a folder before zipping).
Message was edited by: SteveKir Added BTW

Unzipping .gz files

Hi all,
I've never done any compression-decompression work with Java. I've got some help as to how to unzip .zip files, but am unable to unzip .gz files. Does anyone know how to accomplish this?
All help is appreciated!

I've never done any compression-decompression work
with Java. I've got some help as to how to unzip .zip
files, but am unable to unzip .gz files. Does anyone
know how to accomplish this?Implement it yourself or Google for a lib. JSE can only handle ZIPs.

Auto unzip "zipped" files inside OWB. possible?

Similar Messages

Maybe you are looking for