Error 2 - No such file or directory on a zip file

Hey guys, so I borrowed my friends FreeAgent GoFlex drive to get videos from his Go Pro that he put on there. The files were zip files, and when I click on them to open, the Utility comes up and starts unzipping the file, but when the file is done unzipping, it gets the Error 2 - no such file or directory. I thought maybe I'd put the file onto my desktop and unzip it from there, but same error message comes up.
I just want to know what's going on and are the files corrupted when my friend put it onto his external drive, and if there was a way to access the files since I really want them!
Thanks everyone for helping one and taking a look!

I'm at a loss as to why the files are numbered like that - unless, perhaps, they were zipped by StuffIt Deluxe and segmented? If you try opening the first, unnumbered, archive, do you get the same error message? Maybe ask if StuffIt Deluxe was used - http://www.stuffit.com/mac-deluxe.html. It's pretty expensive for what it does but I can only think, at this point, that the files were zipped and segmented using something.
Clinton

Similar Messages

Has anybody gotten an error when trying to unzip the Encore Content zips file? How did you fix?

Has anybody gotten an error when trying to unzip the Encore Content zips file? How did you fix?

You may have a corrupt download... did you download again?

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.

Saving files as type in a zip file

This is probably more simple than I think it is, but I'm having trouble reading files back correctly (outside the program) after I've programmatically made the zip file.
I'm trying to zip up selected contents of a database by getting the item, and then writing the contents of the database entry out to the zip file through a byte stream. This all works fine, but all the entries in the zip file are "files" with no extension, so they don't open properly, if at all. Is there a way to save file types correctly in a zip file based on content? The database doesn't preserve filenames, all items are referred to by unique ID's. Any help is much appreciated.

cotton.m wrote:
nsop07 wrote:
cotton.m wrote:
nsop07 wrote:
This is probably more simple than I think it is, but I'm having trouble reading files back correctly (outside the program) after I've programmatically made the zip file.
I'm trying to zip up selected contents of a database by getting the item, and then writing the contents of the database entry out to the zip file through a byte stream. This all works fine, but all the entries in the zip file are "files" with no extension, so they don't open properly, if at all. Is there a way to save file types correctly in a zip file based on content? The database doesn't preserve filenames, all items are referred to by unique ID's. Any help is much appreciated.The bottom line is that your question has nothing to do with zipping at all. You have files and they aren't saved with an extension. I would suggest you get that changed so that when you pull the data from the BLOB filed you can give it a unique name and sensible extension.
That will solve your problem.
The zipping is an extra step but totally irrelevent here.Thanks for the response.
I understand what you're saying, but I can't change the way the information is being stored or pulled from the database, I'm going to be honest with you here this is a big mistake. Fixing what meta data is available about the file in the database is the proper solution here. The person who originally made the database design made a serious error in not accounting for this information.
There is no simple way to do this. You'll have to actually look at the content of the file and determine as best you can what type of file it is. There are some third party libraries that do this sort of thing like http://sourceforge.net/projects/jmimemagic/
But depending on how many types of data this could get ugly in a hurryThat is the OPEN question here though ... how many type of data. If more than say, one, you're so right. But if it's all the same stuff, than there sould be no pblm.

How to get the entries in a jar/zip file contained within a jar/zip file?

If I want to list the jar/zip entries in a jar/zip file contained within a jar/zip file how can I do that?
In order to get to the entry enumeration I need a Zip/JarFile:
ZipFile zip = new ZipFile("C:/java_dev/Java_dl/java_jdk_commander_v36d.zip");
// Process the zip file. Close it when the block is exited.
try {
// Loop through the zip entries and print the name of each one.
for (Enumeration list =zip.entries(); list.hasMoreElements(); ) {
ZipEntry entry = (ZipEntry) list.nextElement();
System.out.println(entry.getName());
finally {
zip.close();
Zip file "java_jdk_commander_v36d.zip" contains two zip entries:
1) UsersGuide.zip
2) JDKcommander.exe
How to list the entries in "jar:file:/C:/java_dev/Java_dl/java_jdk_commander_v36d.zip!/UsersGuide.zip"?
The following code:
URL url = new URL("jar:file:/C:/java_dev/Java_dl/java_jdk_commander_v36d.zip!/UsersGuide.zip");
JarURLConnection jarConnection = (JarURLConnection)url.openConnection();
zipFile = (ZipFile)jarConnection.getJarFile();
would point to "jar:file:/C:/java_dev/Java_dl/java_jdk_commander_v36d.zip", which is no help at all and Class JarURLConnection does not have an enumeration method.
How can I do this?
Thanks.
Andre

I'm not sure I understand the problem. The difference between a zip and jar file is the manifest file; JarFile is extended from ZipFile and is able to read the manifest, other than that they are the same. Your code
for (Enumeration list =zip.entries(); list.hasMoreElements(); ) {
ZipEntry entry = (ZipEntry) list.nextElement();
System.out.println(entry.getName());
}is close to what I've use for a jar, below. Why not use the same approach? I don't understand what you're trying to do by using JarURLConnection - it's usually used to read the jar contents.
String jarName = "";
JarFile jar = null;
try
    jar = new JarFile(jarName);
catch (IOException ex)
    System.out.println("Unable to open jarfile" + jarName);
    ex.printStackTrace();
for ( Enumeration en = jar.entries() ; en.hasMoreElements() ;)
    System.out.println(en.nextElement());
}

Can't print a PDF file that is within a ZIP file.

Can’t print a PDF file that is within a ZIP file. Get message “Windows cannot complete the transaction. The destination file could not be created.” If I save the file and extract the pdf it works. I didn't have to go through this hastle before. If I retry many times it will often work.

ZIP files can be handled in many different ways. If you have a program like WinZip or 7-Zip installed, they will handle it.
If not, Windows will handle it. The message you quote indicates that your situation is handled by Windows, which seems to have a problem with it. Could be caused by recent Windows updates.
Definitely not an Adobe Reader problem.

No luck trying to extract .inf files from drivers downloaded as zip files

Hi,
I am trying to extract .inf files from drivers downloaded as zip files from the Toshiba website, but I'm having no luck.
I am trying to deploy images using WDS on server 2008 and need the inf files for this to succeed, does anyone have the answer to this problem.
thanks in advance.

Hi
I found this tip:
Someone wrote this:
+You have to run the installer, wait for it to get to the first point where you click Next to continue, and then leave it sitting there. Go into your system's Temp and find the particular one with the installation files inside it that were extracted, then copy it someplace, go back to the installer and cancel it.+
You could also try this freeware application to extract the package:
http://www.legroom.net/software/uniextract

Error 2 - No such file or directory issue with unzipping file

Hi all,
I know ther's been a couple of threads about this already and I can promise you that I have tried it all.
Keep getting Error 2 - No such file or directory
I've downloaded various different software and tried unzipping, tried through Terminal, all of it.
Still, 3 out of 16 files that I downloaded from the same source appear to be corrupted.
Please let me know if anyone of you had this problem and how you solved it?
Thanks a lot

I'm at a loss as to why the files are numbered like that - unless, perhaps, they were zipped by StuffIt Deluxe and segmented? If you try opening the first, unnumbered, archive, do you get the same error message? Maybe ask if StuffIt Deluxe was used - http://www.stuffit.com/mac-deluxe.html. It's pretty expensive for what it does but I can only think, at this point, that the files were zipped and segmented using something.
Clinton

Problem exporting '.txt' file size 23 KB and '.zip' file size 4 MB

I am using Apex 3.0 version screen to upload '.txt' file and '.zip' file containing images.
I can successfully export '.txt' file and '.zip' file containing images as long as '.txt' file size is < 23 KB and '.zip' file size < 4 MB from database table 'TBL_upload_file' to the OS directory on the server.
processing of Larger files (sizes 35 KB and 6 MB) produce following Error Message.
‘ORA-21560: argument 2 is null, invalid or out of range’ error.
Here is my code:
I am using following code to export Documents from database table 'TBL_upload_file' to the OS directory on the server.
create or replace procedure "PROC_LOAD_FILES_TO_FLDR_BYTES"
(pchr_text_file IN VARCHAR2,
pchr_zip_file IN VARCHAR2)
is
lzipfile varchar(100);
lzipname varchar(100);
sseq varchar(1000);
ldocname varchar(100);
lfile varchar(100);
-- loaddoc (p_file in number) as
l_file UTL_FILE.FILE_TYPE;
l_buffer RAW(32000);
l_amount NUMBER := 32000;
l_pos NUMBER := 1;
l_blob BLOB;
l_blob_len NUMBER;
l_file_name varchar(200);
l_doc_name varchar(200);
a_file_name varchar (200);
end_pos NUMBER;
begin
-- Get LOB locator
SELECT blob_content,doc_name
INTO l_blob,l_file_name
FROM tbl_upload_file
WHERE DOC_NAME = pchr_text_file;
--get length of blob
l_blob_len := DBMS_LOB.getlength(l_blob);
-- save blob length to determine end position
end_pos:= l_blob_len;
-- Open the destination file.
-- l_file := UTL_FILE.fopen('BLOBS','MyImage.gif','w', 32767);
l_file := UTL_FILE.fopen('BLOBS',l_file_name,'WB', 32760); --use write byte option supported in 10G
-- if small enough for a single write
IF l_blob_len < 32760 THEN
utl_file.put_raw(l_file,l_blob);
utl_file.fflush(l_file);
ELSE -- write in pieces
-- Read chunks of the BLOB and write them to the file
-- until complete.
WHILE l_pos < l_blob_len LOOP
DBMS_LOB.read(l_blob, l_amount, l_pos, l_buffer);
UTL_FILE.put_raw(l_file, l_buffer);
utl_file.fflush(l_file); --flush pending data and write to the file
-- set the start position for the next cut
l_pos := l_pos + l_amount;
-- set the end position if less than 32000 bytes, here end_pos captures length of the document
end_pos := end_pos - l_amount;
IF end_pos < 32000 THEN
l_amount := end_pos;
END IF;
END LOOP;
END IF;
--- zip file
-- Get LOB locator to locate zip file
SELECT blob_content,doc_name
INTO l_blob,l_doc_name
FROM tbl_upload_file
WHERE DOC_NAME = pchr_zip_file;
l_blob_len := DBMS_LOB.getlength(l_blob);
-- save blob length to determine end position
end_pos:= l_blob_len;
-- Open the destination file.
-- l_file := UTL_FILE.fopen('BLOBS','MyImage.gif','w', 32767);
l_file := UTL_FILE.fopen('BLOBS',l_doc_name,'WB', 32760); --use write byte option supported in 10G
-- if small enough for a single write
IF l_blob_len < 32760 THEN
utl_file.put_raw(l_file,l_blob);
utl_file.fflush(l_file); --flush out pending data to the file
ELSE -- write in pieces
-- Read chunks of the BLOB and write them to the file
-- until complete.
l_pos:=1;
WHILE l_pos < l_blob_len LOOP
DBMS_LOB.read(l_blob, l_amount, l_pos, l_buffer);
UTL_FILE.put_raw(l_file, l_buffer);
UTL_FILE.fflush(l_file); --flush pending data and write to the file
l_pos := l_pos + l_amount;
-- set the end position if less than 32000 bytes, here end_pos contains length of the document
end_pos := end_pos - l_amount;
IF end_pos < 32000 THEN
l_amount := end_pos;
END IF;
END LOOP;
END IF;
-- Close the file.
IF UTL_FILE.is_open(l_file) THEN
UTL_FILE.fclose(l_file);
END IF;
exception
WHEN NO_DATA_FOUND THEN
RAISE_APPLICATION_ERROR(-20214,'Screen fields cannot be blank, Proc_Load_Files_To_Fldr_BYTES.');
WHEN TOO_MANY_ROWS THEN
RAISE_APPLICATION_ERROR(-20215,'More than one record exist in the tbl_load_file table, Proc_Load_Files_To_Fldr_BYTES.');
WHEN OTHERS THEN
-- Close the file if something goes wrong.
IF UTL_FILE.is_open(l_file) THEN
UTL_FILE.fclose(l_file);
END IF;
RAISE_APPLICATION_ERROR(-20216,'Some other errors occurred, Proc_Load_Files_To_Fldr_BYTES.');
end;
I am new to the Oracle.
Any help to modify this scipt and resolve this problem will be greatly appreciated.
Thank you.

Ask this question in the Apex forums. See Oracle Application Express (APEX)
Regards Nigel

Excel 2013 (office 365) crashes every time I opened a excel file directly from Browser or zip file.

Hi guys,
I got a problem with my Office 365 that every time I open an office file from IE/chrome browser (i.e. click the document link and save the file to local, open it in the browser after download completed) or under a zip file, no matter it is an excel file
or word document, the Excel/Word crashes. The Excel/Word will ask me to repair the file if I open the file again. The file will be closed automatically after the repair, when I open the file again it will told me the excel/word itself has just successfully
registered and ask me to close and re-open the excel again in order to enable editing the file. This is extremely inconvenience, I have to open this file 4 times to edit my file.
Could anyone here can help me to resolve the problem?
Many thanks.

Hi,
According to your description, Excel seems not crash when you opened a local Excel/Word file directly. Where do you store these file? OneDrive, SharePoint or other?
If possible, please let me see the screenshot of the error dialog.
Moreover, please open Event Viewer to check whether there's any error about the Excel/Word crash issue.
Press Win + R, type "eventvwr" in the blank box, press Enter. In the Event Viewer, browse to Windows Logs -> Application, there may be some errors after the crash.
To protect your privacy, you may send the error logs to
[email protected], we will try to find the cause.
Regards,
George Zhao
Forum Support
Come back and mark the replies as answers if they help and unmark them if they provide no help.
If you have any feedback on our support, please click "[email protected]"

Converting html file into zip file and send email attaching zip file

Hi Experts,
I am trying to send email with attachment(html). Which contains more than 7MB. So, It is throwing an error like Size exceeded.
So, Now i need to compress the data for less than 7MB.
I decided to convert HTML File into ZIP File.
Kindly suggest me to convert the HTML file into ZIP file and sending email with attached ZIP file.
Correct answer rewarded,
Thanks & Regards,
N. HARISH KUMAR

Hi Experts,
*// HTML_TAB converting into ZIP File
   DATA : zip_tool TYPE REF TO cl_abap_zip,
           filename TYPE string ,
           filename_zip TYPE string .
   DATA : t_data_tab TYPE TABLE OF x255,
           bin_size TYPE i,
           buffer_x TYPE xstring,
           buffer_zip TYPE xstring.
filename = text-007.                                                                          "'HTML_TAB
*describe the attachment
   DESCRIBE TABLE html_tab LINES tab_lines.
   bin_size = tab_lines * 255.
   CALL FUNCTION 'SCMS_BINARY_TO_XSTRING'
     EXPORTING
       input_length = bin_size
     IMPORTING
       buffer       = buffer_x
     TABLES
       binary_tab   = html_tab.
   IF sy-subrc <> 0.
*     message id sy-msgid type sy-msgty number sy-msgno
*     with sy-msgv1 sy-msgv2 sy-msgv3 sy-msgv4.
   ENDIF.
*create zip tool
   CREATE OBJECT zip_tool.
*add binary file
   CALL METHOD zip_tool->add
     EXPORTING
       name    = 'FSSAI_MAIL.HTML'
       content = buffer_x.
*get binary ZIP file
   CALL METHOD zip_tool->save
     RECEIVING
       zip = buffer_zip.
   CLEAR: t_data_tab[],bin_size.
   CALL FUNCTION 'SCMS_XSTRING_TO_BINARY'
     EXPORTING
       buffer        = buffer_zip
     IMPORTING
       output_length = bin_size
     TABLES
       binary_tab    = html_tab.
Thanks & Regards,
N. HARISH KUMAR

Download files from document library in zip file using javascript

Hi,
I am developing SharePoint Hosted App which needs to download all the files in the document libraries of SharePoint site in a zip folder where in that zip file each document library should be a folder, in that folder the files of that document library will
be shown. Currently I am display all the document library in a table with check boxes as shown in the below figure. Now when user selects the check boxes and click on download button, the selected document libraries along with the files should be downloaded
in a zip folder. This should be achieved through javascript. Can anyone share me the code to solve this problem.
Thanks & Regards,
Kishore
Kishore

You can use
zip.js, but the best way - implement your solution on the server-side.
[custom.development]

Xml file to mail scenario with zip file as an attachment

Hi experts,
I have a file to mail scenario. File is in xml format.
At receiver side, first I want to zip this file and send the zip file as an attachment using mail adapter.
How can i achieve this?
Regards
Divia

Hi Shabarish,
In the module tab i have specified the below beans
localejbs/AF_Modules/MessageTransformBean                           contentType
AF_Modules/PayloadZipBean                                                    zip
sap.com/com.sap.aii.adapter.mail.app/XIMailAdapterBean           mail
In the module configuration i mentioned as
Transform.ContentDescription   file
Transform.ContentDisposition   attachment:filename="file.xml"
zip.filenameKey                      contentType
zip.mode                                zipOne
Now i am getting the mail with zip file as an attachment.But the name of the attachment i got is MainDocument.zip
Even the file name inside the zip is MainDocument.xml.
How can i specify my own file name for both zip file and the file inside the archieve folder.Please help me.
Regards
Divia

File to mail scenario with zip file as an attachment

Hi experts,
I have a file to mail scenario.The mail contains a zip file as an attachment.
The file in the input folder is a zip file. I want to send this zip file via email using mail adapter.How can i achieve this?Please help me to solve it out.
Regards
Divia

Hi Narath,
Do check the below link it will be helpfull
Troubleshooting SOAP, HTTP and Mail Adapter Scenarios with TCPGateway
Thanks
Sampath

File name character encoding in zipped file

Hi All,
I have the following problem: I have received a zipped file which contains a folder and file structure. The file was zipped on Windows XP with Czech language. The problem is that the folder/file names use Czech characters (codepage CP1250). The characters in filenames after unzipping the structure are corrupted.
Is there any way how to define the codepage during unzipping or, more in general, way how to convert file names in one codepage to UTF-16 used in Mac's file system?
Thanks for ideas.

You might give this a try for converting the filenames after unzipping, though I think it is designed for the old Mac CE encoding rather than the Windows version:
http://support.apple.com/downloads/FileName_Encoding_RepairUtility
Otherwise the only option may be manually converting the text strings with something like TextWrangler.

Error 2 - No such file or directory on a zip file

Similar Messages

Maybe you are looking for