Pickup and process zip file in watch folder

Hi,
Could someone refer me to an example where it shows me how to use Java script in the Execute Script service to do the following:
- Pickup the zip file from the Watch Folder
- Unzip the zip file
- Loop through all the folders, look for the documents and store them in the list document variable in the process
Thank you in advance for the help!
Hai

You can certainly remove that file. it's basically the cached applet demo.jar (the JRE caches the downloaded jars so that you don't have to keep re-downloading them each time you visit the web site).
The fact that your Antivirus complains about it just means that it's confused. We're all using the Sun environment, and I have the latest and greatest Norton, and there are no complaints, so it's clear that your Antivirus software needs a fix.

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We use to create watched folder for every project in Distiller. A single distiller is having more than 30 watched folder.
Just wanted to check how distiller watched folder works:
1. Is the distiller pickup PS file as alphabetical wise?
2. Is the distiller pickup PS file as priority when hitting the folder?
3. Is the distiller pickup PS file as time wise?
We had an experience the PS file has picked up as alphabetical wise? Is this right?
I need to understand whether the watched folder should pickup the file based on the time?
Kindly clarify.
JA

I'm assuming this isn't the discontinued Distiller Server product...
Distiller is for low volume personal (single user) use, so Adobe have no documentation and no interest in tuning the performance of multiple queues.
Since each end user MUST have an Acrobat license you might as well set them up individual queues on their own PCs.

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Hi Chris,
- Yes. PDFG was installed on machine having an Office2003 install.
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Saved presentation as .zip file or a folder. Presentation gone?

Hello, I really need some help. Yesterday, I worked with a presentation which took about 10 hours, I'm not really that experienced with computers, but I think I saved it as a .zip file or simply as a folder. I cannot find the presentation and open it anymore, and the .zip file and the folder simply contains:
All the images i used in the presentation.
A folder named "Quicklook".
A folder named "Thumbnails", with all the thumbnails of each slide, but they are too small to read, and it doesn't help to zoom in, too bad quality.
The frontpage of my presentation.
A file named "Index", not sure about the filetype.
I really need help on how to fix this, if it is possible. I don't understand where the main file for the presentation is, it seems like this folder only contains all the materials and parts for the presentation.
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Thanks in advance.

Welcome to the discussions, Maryam222.
Keynote "files" are actually .zip files or folders with the difference being that their name has .key on the end. As long as you haven't moved any of the items/objects out of OR to different positions in that folder, you should be able to...
1) Select the icon that you think is your presentation (not the stuff on the inside, the folder that it's all in, select the folder)
2) Do "command-I" to "Get Info" for that icon.
3) In the "Name & Extension" area, you should see the name of the folder
4) I'm guessing that your problem is that it doesn't have .key at the end. So, add .key to the end of the filename (If it has .zip at the end, change .zip to .key)
5) OSX will tell you something about changing the file, click the button to confirm the change
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Zipping files in a folder using java.util.zip

hi guys,
does any body has the java code for
zipping files in a folder
hi guys,
actually i want to zip no of files in a folder .the folder can also have sub folder
and after zipping when i extract the files manually i should
maintain the same folder structure
does any body has the code
please reply me soon its a bit urgent

hi smeee
i tried running ur code but its not working
here the command
java c:\abc\ zzz.zip
its saying
no files or directories even though there's the
directory present
what's the solution Hi,
Oops that was because of the error check added at the last moment..
Anyway, You need to use the following command to run it :
java ZipUtility c:\abc c:\zzz.zip
Please use the following corrected code:
 Class to zip file(s).
 Requires jdk1.3.1.
 To run use the following command:
 java ZipUtility <directory or file to be zipped> <name of zip file to be created>
 Please specify absolute path names as its parameters while running this program.
import java.util.zip.*;
import java.io.*;
public class ZipUtility {
 ZipOutputStream cpZipOutputStream = null;
 String strSource = "";
 String strTarget = "";
 String strSubstring = "";
 public static void main(String args[]) {
 if( args == null || args.length < 2) {
 System.out.println("Usage: java ZipUtility <directory or file to be zipped> <name of zip file to be created>");
 return;
 ZipUtility udZipUtility = new ZipUtility();
 udZipUtility.strSource = args[0];
 udZipUtility.strTarget = args[1];
 udZipUtility.zip();
 private void zip(){
 try
 File cpFile = new File (strSource);
 if (!cpFile.isFile() && !cpFile.isDirectory() ) {
 System.out.println("\nSource file/directory Not Found!");
 return;
 if (cpFile.isDirectory()) {
 strSubstring = strSource;;
 } else {
 strSubstring = "";
 cpZipOutputStream = new ZipOutputStream(new FileOutputStream(strTarget));
 cpZipOutputStream.setLevel(9);
 zipFiles( cpFile);
 cpZipOutputStream.finish();
 cpZipOutputStream.close();
 System.out.println("\n Finished creating zip file " + strTarget + " from source " + strSource);
 }catch (Exception e){
 e.printStackTrace();
 private void zipFiles(File cpFile) {
 if (cpFile.isDirectory()) {
 File [] fList = cpFile.listFiles() ;
 for (int i=0; i< fList.length; i++){
 zipFiles(fList) ;
 } else {
 try {
 String strAbsPath = cpFile.getAbsolutePath();
 String strZipEntryName ="";
 if (!strSubstring.equals("") ){
 strZipEntryName = strAbsPath.substring(strSource.length()+1, strAbsPath.length());
 } else {
 strZipEntryName = cpFile.getName();
 byte[] b = new byte[ (int)(cpFile.length()) ];
 FileInputStream cpFileInputStream = new FileInputStream (cpFile) ;
 int i = cpFileInputStream.read(b, 0, (int)cpFile.length());
 ZipEntry cpZipEntry = new ZipEntry(strZipEntryName);
 cpZipOutputStream.putNextEntry(cpZipEntry );
 cpZipOutputStream.write(b, 0, (int)cpFile.length());
 cpZipOutputStream.closeEntry() ;
 } catch (Exception e) {
 e.printStackTrace();

Create Zip File In Windows and Extract Zip File In Linux

I had created a zip file (together with directory) under Windows as follow (Code are picked from [http://www.exampledepot.com/egs/java.util.zip/CreateZip.html|http://www.exampledepot.com/egs/java.util.zip/CreateZip.html] ) :
package sandbox;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.zip.ZipEntry;
import java.util.zip.ZipOutputStream;
* @author yan-cheng.cheok
public class Main {
 * @param args the command line arguments
 public static void main(String[] args) {
 // These are the files to include in the ZIP file
 String[] filenames = new String[]{"MyDirectory" + File.separator + "MyFile.txt"};
 // Create a buffer for reading the files
 byte[] buf = new byte[1024];
 try {
 // Create the ZIP file
 String outFilename = "outfile.zip";
 ZipOutputStream out = new ZipOutputStream(new FileOutputStream(outFilename));
 // Compress the files
 for (int i=0; i<filenames.length; i++) {
 FileInputStream in = new FileInputStream(filenames);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(filenames[i]));
// Transfer bytes from the file to the ZIP file
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
// Complete the entry
out.closeEntry();
in.close();
// Complete the ZIP file
out.close();
} catch (IOException e) {
e.printStackTrace();
The newly created zip file can be extracted without problem under Windows, by using [http://www.exampledepot.com/egs/java.util.zip/GetZip.html|http://www.exampledepot.com/egs/java.util.zip/GetZip.html]
However, I realize if I extract the newly created zip file under Linux, using modified version of [http://www.exampledepot.com/egs/java.util.zip/GetZip.html|http://www.exampledepot.com/egs/java.util.zip/GetZip.html] . The original version doesn't check for directory using zipEntry.isDirectory()).public static boolean extractZipFile(File zipFilePath, boolean overwrite) {
InputStream inputStream = null;
ZipInputStream zipInputStream = null;
boolean status = true;
try {
inputStream = new FileInputStream(zipFilePath);
zipInputStream = new ZipInputStream(inputStream);
final byte[] data = new byte[1024];
while (true) {
ZipEntry zipEntry = null;
FileOutputStream outputStream = null;
try {
zipEntry = zipInputStream.getNextEntry();
if (zipEntry == null) break;
final String destination = Utils.getUserDataDirectory() + zipEntry.getName();
if (overwrite == false) {
if (Utils.isFileOrDirectoryExist(destination)) continue;
if (zipEntry.isDirectory())
Utils.createCompleteDirectoryHierarchyIfDoesNotExist(destination);
else
final File file = new File(destination);
// Ensure directory is there before we write the file.
Utils.createCompleteDirectoryHierarchyIfDoesNotExist(file.getParentFile());
int size = zipInputStream.read(data);
if (size > 0) {
outputStream = new FileOutputStream(destination);
do {
outputStream.write(data, 0, size);
size = zipInputStream.read(data);
} while(size >= 0);
catch (IOException exp) {
log.error(null, exp);
status = false;
break;
finally {
if (outputStream != null) {
try {
outputStream.close();
catch (IOException exp) {
log.error(null, exp);
break;
if (zipInputStream != null) {
try {
zipInputStream.closeEntry();
catch (IOException exp) {
log.error(null, exp);
break;
} // while(true)
catch (IOException exp) {
log.error(null, exp);
status = false;
finally {
if (zipInputStream != null) {
try {
zipInputStream.close();
} catch (IOException ex) {
log.error(null, ex);
if (inputStream != null) {
try {
inputStream.close();
} catch (IOException ex) {
log.error(null, ex);
return status;
*"MyDirectory\MyFile.txt" instead of MyFile.txt being placed under folder MyDirectory.*
I try to solve the problem by changing the zip file creation code to
+String[] filenames = new String[]{"MyDirectory" + "/" + "MyFile.txt"};+
But, is this an eligible solution, by hard-coded the seperator? Will it work under Mac OS? (I do not have a Mac to try out)
p/s To be honest, I do a cross post at [http://stackoverflow.com/questions/2549766/create-zip-file-in-windows-and-extract-zip-file-in-linux|http://stackoverflow.com/questions/2549766/create-zip-file-in-windows-and-extract-zip-file-in-linux] Just want to get more opinion on this.
Edited by: yccheok on Apr 26, 2010 11:41 AM

Your solution lies in the File.separator constant; this constant will contain the path separator that is used by the operating system. No need to hardcode one, Java already has it.
edit: when it comes to paths by the way, I have the bad habit of always using the front slash ( / ). This will also work under Windows and has the added benefit of not needing to be escaped.

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Error message(s): None.
Error present in specific output format/encoding preset? No. It's a global issue.
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Has this ever worked before? Yes, in CS5.5 Production Bundle's AME.
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thanks

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Hi,
I have a scenario where we need to pick up a file from a folder location on a particular date of a month and send it via email to an email id.
For this, we have planned a design where I have a SQL receive adapter which will poll a table in SQL and return a boolean value indicating that the file is
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once the orchestration instance is created, in the next steps, I am trying to receive the file from a folder location. I am getting a build error here. Mentioning "you must specify at least one already-initialized correlation set for a non-activation receive
that is on a non-selfcorrelating port".
Can you please suggest how I can correlate on these locations.
Regards,
Ujjwal
-Ujjwal

Ok got,
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SQL to receive the trigger message and in second Receive shape you expect to receive the message from Folder (file location)
since you have two receive location in your Orchestration you have a convoy and you have set correlation for your Receive shapes. Initialize the correlation set
in first Receive shape and follow the correlation set in another Receive shape.
But for your requirement, even when the flow of your Orchestration moves to the second Receive shape (which is configured to
file location) it not going to pull/receive the file into your Orchestration as you expect.
As said, read the
schedule task adapter and use to trigger the file receive process on particular date of a month and have send
port with filter for the Receive port name (which has Receive location with schedule task adapter) and configure this send port to send email.
Read more about correlation here:
Using Correlations in Orchestrations
Walkthrough: Correlations in BizTalk Orchestration
Read more about convoy here.
Working with Convoy Scenarios
Sequential Convoys
Read more about
Schedule Task Adapter:
https://biztalkscheduledtask.codeplex.com/
https://connectedenterprise.wordpress.com/2011/09/26/the-very-useful-biztalk-scheduled-task-adapter/
If this answers your question please mark it accordingly. If this post is helpful, please vote as helpful by clicking the upward arrow mark next to my reply.

Upload and Process large files

We have a SharePoint 2013 OnPrem installation and have a business application that provides an option to copy local files into UNC path and some processing logic applied before copying it into SharePoint library. The current implementation is
1. Users opens the application and clicks “Web Upload” link from left navigation. This will open a \Layouts custom page to select upload file and its properties
2. User specifies the file details and chooses a Web Zip file from his local machine
3. Web Upload Page Submit Action will
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     b. Creates a list item to store its properties along with the UNC path details
4. Timer Job executes in a periodic interval to
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Can someone suggest a different design approach that can manage the large file outside of SharePoint context? Something like
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   2. Instead of timer jobs, have external services that grab data from a UNC path and processes periodic intervals to push it into SharePoint.

Hi,
According to your post, my understanding is that you want to upload and process files for SharePoint 2013 server.
The following suggestion for your reference:
1.We can create a service to process the upload file and copy the files to the UNC folder.
2.Create a upload file visual web part and call the process file service.
Thanks,
Dennis Guo
TechNet Community Support
Please remember to mark the replies as answers if they help and unmark them if they provide no help. If you have feedback for TechNet Subscriber Support, contact
[email protected]
Dennis Guo
TechNet Community Support

Pickup and process zip file in watch folder

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