Problems creating/extracting ZIP files, anyone?

Greetings,
I am having some odd bugs with OSX in creating and extracting ZIP files. I’m curious if anyone else has run across this or if anyone has any remedies to suggest.
The issue is that when I extract a ZIP archive, the extracted files have their creation dates and modification dates changed from the original versions. For example, I have a directory created in 2010 and all the files were created and last modified in 2010. However, when I ZIP the directory and re-extract it, many of the file dates change. Some files have the creation dates changed so to match the last modified date. In other cases, both the creation date and modification date are changed to something later than both, perhaps the date the directory was ZIPed or the date the directory was extracted. And some files don’t seem to have the dates changed.
I have tried various options, such as terminal, and third party programs (BetterZip, Keka, Stuffit, & Archiver to name a few) and they all seem to have some variation of this behavior. I have tried creating the Zip with one program and extracting with the other, and still this behavior persists. Different files seem to be affected with each combination of creating and extracting programs.
Another odd behavior was that I tried password protecting some directories with financial data, and OSX could create the ZIP file with the password, but OSX could not unzip the file with the password it created for some files. In every case, third party program could unzip the files with the password created in the terminal command. This seem to happen inconsistently.
I have tried creating a dummy account and tested it in that account, with similar results.
I’ve read the man page for the command line version of ZIP to see if there is some -option to preserve dates, but found nothing. The typical command looks like this:
> zip -er filename.zip sourceDir
> unzip filename.zip
File types include directories, PDFs, Text docs, Quicken files. Ive tested other directories too with other file types. It doesn’t seem to affect every file, but seems unpredictable. Luckily, I still have the original uncompressed version of most directories. But I really want to archive some directories and unarchive them when needed. The dates matter to me because they are mostly financial files and I use the dates to know what versions were used for taxes and other things. Also, if dates change, then backup programs start replacing the good backups with files with the corrupted dates.
I’d appreciate any suggestions or feedback. I there some bug in the ZIP protocol? Is there some bug in OSX file system that is corrupting the dates with the archive and unarchive process?
Thank you for any suggestions-
OSX 10.9.5,
MBPro 2012

OK, sorry did not read it correctly. Read the first sentance and quit.
No, Bridge is just a browser. You can extract more than one file at a time in Win. Explorer. You have to do them individually, but can run multiple extractions at the same time.
There are other zip/unzip programs out there. I know the the CS6 beta a number of users could not open it with WinZip and used alternate product just fine.

Similar Messages

Having problems creating a zip file with Japanese language character names

I have a bunch of files with names in Japanese characters (also Chinese, Korean, Spanish etc, but this will do for an example). The encoding is Unicode.
I wish to be able to put them in a zip file, then extract them again using any old zip tool (WinZip,PKZip,7-Zip etc)
Trouble is, every time I do it, the files inside the zipfile end up with garbage character names. The contents of the files are fine, though.
I'm aware that there used to be a bug in the java.util.zip.* classes regarding character encodings (http://bugs.sun.com/bugdatabase/view_bug.do;jsessionid=5bd4fe01ad8a7b4ec89afef5005da?bug_id=4244499) but as far as I can see it's supposed to have been fixed I have also tried the ZipOutputStream class from apache which allows you to set the encoding manually - no luck there either (just many different varieties of garbage characters)
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import java.io.File;
import java.io.FileOutputStream;
import java.io.FileReader;
import java.io.IOException;
public class ZipTest {
static String uniqueFileName = "西　純.txt";
public static void main(String[] args) {
try {
standardZip();
apacheZip();
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String encoding = "";
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encoding = reader.getEncoding();
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org.apache.tools.zip.ZipOutputStream zipOutputStream = new org.apache.tools.zip.ZipOutputStream(
zipFile);
zipOutputStream.setEncoding(encoding);
org.apache.tools.zip.ZipEntry zipEntry = new org.apache.tools.zip.ZipEntry(uniqueFileName);
zipEntry.setSize(inputFile.length());
zipEntry.setTime(inputFile.lastModified());
zipOutputStream.putNextEntry(zipEntry);
int c = 0;
while (c >= 0) {
c = reader.read();
if (c >= 0) {
zipOutputStream.write(c);
zipOutputStream.closeEntry();
zipOutputStream.finish();
public static void standardZip() throws IOException {
String encoding = "";
File inputFile = new File(uniqueFileName);
FileReader reader = new FileReader(inputFile);
encoding = reader.getEncoding();
System.out.println("Using Java IO - Encoding = " + encoding);
File zipFile = new File("standardZip.zip");
FileOutputStream zipOut = new FileOutputStream(zipFile);
java.util.zip.ZipOutputStream zipOutputStream = new java.util.zip.ZipOutputStream(zipOut);
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zipEntry.setSize(inputFile.length());
zipEntry.setTime(inputFile.lastModified());
zipOutputStream.putNextEntry(zipEntry);
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while (c >= 0) {
c = reader.read();
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Emma_Baillie wrote:
I have a bunch of files with names in Japanese characters (also Chinese, Korean, Spanish etc, but this will do for an example). The encoding is Unicode.
I wish to be able to put them in a zip file, then extract them again using any old zip tool (WinZip,PKZip,7-Zip etc)
Trouble is, every time I do it, the files inside the zipfile end up with garbage character names. The contents of the files are fine, though.This is becuase zip tool doesn't support unicode for zip entries. For example WinZip prior to 11.2 does not support Unicode characters in filenames. You need to look for the simillar information for other tools. You can find more on tools on their website.

Problem In Extracting xml file from Zip File

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In my application I am creating a zip file which contains some xml files. I am using Weblogic 8.1 for my application.
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Friends
I resolved the problem
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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 {
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            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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I am currently extracting data from a table and creating a flat file which is read later to create a compressed (zip) file.
Is there a way that I can create a zipped file in one step (i.e. directly from the table data)? I am trying to reduce the number of file i/o hits in my appication.
Thanks,
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Dear all,
Following my yesterday's posting, I would clarify my questions:
I successfully created a zip file which contains a folder structure and two files from strings. See attached codes.
However, I don't like the way that I physically created subfolder and files before add them to the zip file.
Is there any possibility to avoid that part of creating multiple folders?
Regards.
Pengyou
package com.zip.test;
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.FileWriter;
import java.io.IOException;
import java.util.zip.ZipEntry;
import java.util.zip.ZipOutputStream;
public class ZipTest {
     static final int BUFFER = 2048;
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                         System.out.println("Directories: " + strManyDirectories
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                    File outputFile1 = new File("root/dir1/dir2/test1.html");
                    BufferedWriter bufferedWriter1 = new BufferedWriter(
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               } catch (IOException e) {
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pengyou wrote:
But I did not understand how to convert a String to ZipOutputStream with a user choosen file name and a user choosen folder structure.
I am blocked at that point. Sorry for repeating the same question.
That doesn't make sense to me. Even more so as your sample code suggests that you are familiar with Writer and ZipEntry concepts. Can you explain in detail where you are stuck or what part of my previous suggestion was unclear to you?
Scratch that, I think I finally understand your problem. I think the following sample code will get you on track:import java.util.zip.ZipOutputStream;
import java.util.zip.ZipEntry;
import java.io.FileOutputStream;
import java.io.BufferedWriter;
import java.io.OutputStreamWriter;
public class ZipTest {
    public static void main( String[] args ) throws Exception {
        final String[] content = {"bla", "blubber"};
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        final String path = System.getProperty("user.home") + "/test.zip";
        ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(path));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(zos, "UTF-8"));
        for (int i = 0; i < content.length; ++i) {
            final ZipEntry entry = new ZipEntry(names);
zos.putNextEntry(entry);
bw.write(content[i]);
bw.flush();
zos.closeEntry();
// this will close zos, too
bw.close();
Edited by: thomas.behr on Apr 22, 2009 4:10 PM

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I need some quick response on it.

Nothing to do with the resource path....thats fine. Use the code below.....most likely this will fix it.......
Added a check for directories that start with a ".". The File utility probably thinks its a file, also added a try....check it does make the "." directories.........dont forget the dukes.
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      File ff = null;
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        ff = new File( "C:\\your_path\\" + ze.getName() );
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Regards,
Antonio
[email protected]

With Core Java, for all I know you still can't.

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I need verification from you that after extracting zip file I have total size 38 gb after extracting zip file of oracle R12.1.1 for solaris 5.1 sparc 64 bit.
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Thanks

Hi,
I am successfull in putting all the files of a
directory in a newly created ZIP file. But if my
directory has a sub-directory, it's not workin....m
not able to zip it properly. I mean, my program is
doin it withour error but when i try to open it using
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archive...
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I have a new laptop running Windows 7 64 bit but there is no issue opening the pdf in that same zip file received.
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testing has revealed that this has something to do with the rename command or java being unsynchronized with the data actually on the disk (caching problem?).
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          File toErase = new File( ((Integer)lexicon.get(keyword)).intValue()+".zip");
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          // rename the temporary file to the old files name.
          File toRename = new File( ((Integer)lexicon.get(keyword)).intValue()+".tmp.zip");
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we use:
1. j2sdk 1.4.2_05
2. running under libranet linux.
any help will be appreciated

the data does indeed get lost in the course fo rewriting the file, e assume this is some sort of synchronization issue with java and the file system.
addig enteies to an existing zip file would have been great, if we had it a few weeks ago , we had to make do without this ability , and all of the code is already written and mostly working (except for the problem I mention) and we don't really want to rewrite the zip file access.
additional testing has shown that the data gets lost when we use the rename command.
very strange...
thanks for the input

Cannot extract Zip file with Winzip after zipping with java.util.zip

Hi all,
I write a class for zip and unzip the text files together which can be zip and unzip successfully with Java platform. However, I cannot extract the zip file with Winzip or even WinRAR after zipping with Java platform.
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Below is the code:
=====================================================================
package myapp.util;
import java.io.* ;
import java.util.TreeMap ;
import java.util.zip.* ;
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public void createZip(String zipName, String fileName)
      throws ZipException, FileNotFoundException, IOException, UserException {
    FileOutputStream fos = null ;
    BufferedOutputStream bos = null ;
    ZipOutputStream zos = null ;
    File file = null ;
    try {
      file = new File(zipName) ; //new zip file
      if (file.isDirectory()) //check if it is a directory
     throw new UserException("Invalid zip file ["+zipName+"]") ;
      if (file.exists() && !file.canWrite()) //check if it is readonly
     throw new UserException("Zip file is ReadOnly ["+zipName+"]") ;
      if (file.exists()) //overwrite the existing file
     file.delete();
      file.createNewFile();
      //instantiate the ZipOutputStream
      fos = new FileOutputStream(file) ;
      bos = new BufferedOutputStream(fos) ;
      zos = new ZipOutputStream(bos) ;
      this.writeZipFileEntry(zos, fileName); //call to write the file into the zip
      zos.finish() ;
    catch (ZipException ze) {
      throw ze ;
    catch (FileNotFoundException fnfe) {
      throw fnfe ;
    catch (IOException ioe) {
      throw ioe ;
    catch (UserException ue) {
      throw ue ;
    finally {
      //close all the stream and file
      if (fos != null)
     fos.close() ;
      if (bos != null)
     bos.close();
      if (zos != null)
     zos.close();
      if (file != null)
     file = null ;
    }//end of try-catch-finally
private void writeZipFileEntry(ZipOutputStream zos, String fileName)
      throws ZipException, FileNotFoundException, IOException, UserException {
    BufferedInputStream bis = null ;
    File file = null ;
    ZipEntry zentry = null ;
    byte[] bArray = null ;
    try {
      file = new File(fileName) ; //instantiate the file
      if (!file.exists()) //check if the file is not exist
     throw new UserException("No such file ["+fileName+"]") ;
      if (file.isDirectory()) //check if the file is a directory
     throw new UserException("Invalid file ["+fileName+"]") ;
      //instantiate the BufferedInputStream
      bis = new BufferedInputStream(new FileInputStream(file)) ;
      //Get the content of the file and put into the byte[]
      int size = (int) file.length();
      if (size == -1)
     throw new UserException("Cannot determine the file size [" +fileName + "]");
      bArray = new byte[(int) size];
      int rb = 0;
      int chunk = 0;
      while (((int) size - rb) > 0) {
     chunk = bis.read(bArray, rb, (int) size - rb);
     if (chunk == -1)
       break;
     rb += chunk;
      }//end of while (((int)size - rb) > 0)
      //instantiate the CRC32
      CRC32 crc = new CRC32() ;
      crc.update(bArray, 0, size);
      //instantiate the ZipEntry
      zentry = new ZipEntry(fileName) ;
      zentry.setMethod(ZipEntry.STORED) ;
      zentry.setSize(size);
      zentry.setCrc(crc.getValue());
      //write all the info to the ZipOutputStream
      zos.putNextEntry(zentry);
      zos.write(bArray, 0, size);
      zos.closeEntry();
    catch (ZipException ze) {
      throw ze ;
    catch (FileNotFoundException fnfe) {
      throw fnfe ;
    catch (IOException ioe) {
      throw ioe ;
    catch (UserException ue) {
      throw ue ;
    finally {
      //close all the stream and file
      if (bis != null)
     bis.close();
      if (file != null)
     file = null ;
    }//end of try-catch-finally
}

Tried~
The problem is still here~ >___<
Anyway, thanks for information sharing~
The message is:
Cannot open file: it does not appear to be a valid archive.
If you downloaded this file, try downloading the file again.
The problem may be here:
if (fos != null)
fos.close() ;
if (bos != null)
bos.close();
if (zos != null)
zos.close();
if (file != null)
file = null ;
The fos is closed before bos so the last buffer is not
saved.
zos.close() is enough.

Problem creating an jar file!

Hello.
I am trying to create an executable jar file using the following command:
jar cmf MANIFEST.MF game.jar *.classHowever, I keep receiving the following error:
java.io.FileNotFoundException: MANIFEST.MF (The system cannot find the file spec
ified)
        at java.io.FileInputStream.open(Native Method)
        at java.io.FileInputStream.<init>(FileInputStream.java:64)
        at sun.tools.jar.Main.run(Main.java:124)
        at sun.tools.jar.Main.main(Main.java:904)Why am I getting this error and how can I correct this?

Did you create a file called MANIFEST.MF? Is it in the same directory you are in now?
And by the way, according to what I see when I run the jar command, you need to specify the jar file first, and then the manifest file:Usage: jar {ctxu}[vfm0Mi] [jar-file] [manifest-file] [-C dir] files ...
Options:
    -c create new archive
    -t list table of contents for archive
    -x extract named (or all) files from archive
    -u update existing archive
    -v generate verbose output on standard output
    -f specify archive file name
    -m include manifest information from specified manifest file
    -0 store only; use no ZIP compression
    -M do not create a manifest file for the entries
    -i generate index information for the specified jar files
    -C change to the specified directory and include the following file
If any file is a directory then it is processed recursively.
The manifest file name and the archive file name needs to be specified
in the same order the 'm' and 'f' flags are specified.
Example 1: to archive two class files into an archive called classes.jar:
       jar cvf classes.jar Foo.class Bar.class
Example 2: use an existing manifest file 'mymanifest' and archive all the
           files in the foo/ directory into 'classes.jar':
       jar cvfm classes.jar mymanifest -C foo/ .

Problem in deleting Zip files unzipped using java.util.zip

I have a static methos for unzipping a zip file. after unzipping the file when i am trying to delete that file using File.delete()its not getting deleted. but when methods like exist(). canRead(), canWrite() methods are returning true what can be the possible problem ? i had closed all the streams after unzipping operation Please go through the following code.
public static boolean unzipZipFile(String dir_name, String zipFileName) {
try {
ZipFile zip = new ZipFile(zipFileName);
Enumeration entries = zip.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = (ZipEntry) entries.nextElement();
// got all the zip entries here
// now has to process all the files
// first all directories
// then all the files
if (entry.isDirectory()) {
// now the directories are created
File buf=new File(dir_name,entry.getName());
buf.mkdirs();
continue;
}// now got the dirs so process the files
entries = zip.entries();
while(entries.hasMoreElements()) {
// now to process the files
ZipEntry entry = (ZipEntry) entries.nextElement();
if (!entry.isDirectory()){
File buf=new File(dir_name,entry.getName());
copyInputStream(
zip.getInputStream(entry),
new BufferedOutputStream(
new FileOutputStream(buf)));}
} catch (IOException e) {
e.printStackTrace();
return false;
return true;
now i am trying to call this method to unzip a zip file
public static void main (String arg[]){
unzipZipFile("C:/temp","C:/tmp.zip");
java.io.File filer = new File("C:/tmp.zip");
System.out.println (filer.canRead());
System.out.println (filer.canWrite());
System.out.println (filer.delete());
Please tell me where my program is going wrong ?

Thanks .. the problem is solved... i was not closing the Zip file .. rather i was trying to close all the other streams that i used for IO operaion ... thanks a lot

Problems creating/extracting ZIP files, anyone?

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