Get "in memory" zip file's content
Hello,
How can I get the contents of a compressed zip file that is in memory?
Thanks
Karen
The zip file is in memory as a byte array.
However I'm downloading the zip file from a server, and storing it in memory myself so the format is not an issue. I can manipulate it as I need.
Karen
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How do I get a download zip file from firefox to my desktop
I download zip files from different web sites and they go to the download folder in firefox, but I can't access them to open,unzip, or work with them on my computer offline. How can I move them from the download folder in firefox to my computer desktop?
Hi,
In order to open ZIP files, you have to download an app to enable the opening of the ZIP. Here's one you can use: https://play.google.com/store/apps/details?id=com.winzip.android
Next, when you install that, open it and go to your Download directory. It should either be in sdcard0, sdcard, or sdcard1, then go to Download.
Once you find it, touch and hold the file, then click '''Unzip here''' and it will extract the files in the ZIP folder.
Let me know if this works! -
Why do all downloads get downloaded as a zip file?
Recently I was asked about downloading a file as a zipped file. I clicked ok. I didnt pay much attention to it as I thought it was just for that particular file. Now everything I download downloads as a zip file. Anyone know how to fix?
I have not seen this happen. Zip files get saved as Zip files. DMG files get saved as DMG files. In order to convert, and or make, a Zip file out of one or any number of files you need a program, or a feature in the OS, to create a Zip archive file with that one or many files.
Do you have such a program installed? Like Stuffit? -
ByteArrayInputStream as source for zip file entry corrupts the ByteArray
Hi All,
I have managed to create and add a (in-memory) zip file as an email attachment. I add a file to the zip file , then push data to this file via a ByteArrayInputStream.
Adding the file (created via ByteArrayInputStream) to an email works perfectly. If i then add another step and zip this file it corrupts the output.
It looks like it is turning the carriage return into a [] . Its 'almost right' as when I cut & paste the contents of the txt file i created (from the zip file) into Word it displays correctly, with the carriage returns. Is it some kind of encoding problem, or mime type issue?
My ByteArrayInputStream uses buf.append("\n"); for carriage returns. I think the problem is with
ByteArrayInputStream baIn = new ByteArrayInputStream( getAttachementNoFormat(eq_rt.getStoredProc()) );where I think i need to encode this somehow. Here is the complete code to create and add the zip file to an email. Any isuggestions will be appreciated.
//START of new ZIP code
/* Specify files to be zipped */
// Create temp file.
//File temp = File.createTempFile(fileName, ".txt");
//temp.deleteOnExit();
//BufferedWriter bOut = new BufferedWriter(new FileWriter(temp));
//ByteArrayInputStream baInTemp = new ByteArrayInputStream( getAttachementNoFormat(eq_rt.getStoredProc() ) );
//bOut.write( baInTemp.toString() );
//bOut.close();
String[] filesToZip = new String[3];
filesToZip[0] = "C:\\Program Files\\NetBeans3.6\\firstfile.txt";
filesToZip[1] = "C:\\Program Files\\NetBeans3.6\\secondfile.txt";
filesToZip[2] = "C:\\Program Files\\NetBeans3.6\\thirdfile.txt";
final String fileToZip = fileName;
/* Create a memory buffer to store the ByteArray, fixed size */
byte[] buffer = new byte[18024];
/* Specify zip file name */
String zipFileName= eq_rt.getReportName() + ".zip";
try {
/* Create ZipOutputStream to store the FileOutputStream */
// ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipFileName));
ByteArrayOutputStream byteArray = new ByteArrayOutputStream();
ZipOutputStream out = new ZipOutputStream(byteArray);
/* Set the compression ratio */
out.setLevel(Deflater.DEFAULT_COMPRESSION);
/* iterate through the array of files, adding each to the zip file */
for (int a = 0; a < filesToZip.length; a++) {
/* Print the filenumber being added to the zip */
System.out.println(a);
/* Associate a file input stream for the current file */
// FileInputStream in = new FileInputStream(filesToZip[a]);
This ROCKS as it is passing a array into the text file .getBytes() seems
to be the KEY in getting ByteArrayInputStream to WORK
//String strSocketInput = "TAIWAN";
//ByteArrayInputStream baIn = new ByteArrayInputStream(strSocketInput.getBytes());
//String strSocketInput = new String (getAttachementNoFormat(eq_rt.getStoredProc()).toString() );
//ByteArrayInputStream baIn = new ByteArrayInputStream( getAttachementNoFormat(eq_rt.getStoredProc()) );
ByteArrayInputStream baIn = new ByteArrayInputStream( getAttachementNoFormat(eq_rt.getStoredProc()) );
//String strSocketInput = "TAIWAN";
//ByteArrayInputStream baIn = new ByteArrayInputStream(strSocketInput.getBytes());
/* Add ZIP entry to output stream. */
out.putNextEntry(new ZipEntry(filesToZip[a]));
/* Transfer bytes from the current file to the ZIP file */
int len;
while ((len = baIn.read(buffer)) > 0)
out.write(buffer, 0, len);
/* Close the current entry */
out.closeEntry();
/* Close the current file input stream */
baIn.close();
/* Close the ZipOutPutStream (very important to close the zip before you attach it to the email) Thanks DrClap */
out.close();
/* Create a datasource for email attachment */
// DataSource sourcezip = new FileDataSource(zipFileName);
DataSource sourcezip = new ByteArrayDataSource(byteArray.toByteArray(), zipFileName, "application/gzip" );
/* Create a new MIME bodypart */
BodyPart attachment = new MimeBodyPart();
attachment.setDataHandler(new DataHandler(sourcezip));
attachment.setFileName(zipFileName);
/* attach the attachemnts to the mail */
multipart.addBodyPart(attachment);
catch (IllegalArgumentException iae) {
iae.printStackTrace();
catch (FileNotFoundException fnfe) {
fnfe.printStackTrace();
catch (IOException ioe)
ioe.printStackTrace();
// End Of New ZIP codeI came up with 'a' solution (not sure if its the best way but appears to work)
\n = corrupt
\r\n = ok in zip -
I have a corrupted zip file whose contents I'm trying to salvage. I know that the problem is with one of the files in the sbutman/taxes/ directory, and I've resigned myself to losing it. I've written the following short program to try to extract as much data as I can from the archive:
* Created on Sep 7, 2005
import java.io.*;
import java.util.zip.*;
* @author sbutman
public class ZipExtractor {
static final int BUFFER = 8192;
public static void main( String argv[] ) {
try {
BufferedOutputStream dest = null;
FileInputStream fis = new FileInputStream( argv[0] );
ZipInputStream zis = new ZipInputStream( new BufferedInputStream(
fis ) );
ZipEntry entry;
while( true ) {
try {
entry = zis.getNextEntry( );
if ( entry == null ) break;
if ( entry.getName( ).startsWith( "sbutman/taxes/" ) )
continue;
System.out.println( "Extracting: " + entry.getName( ) );
if ( entry.isDirectory( ) ) {
File directoryEntry = new File( entry.getName( ) );
directoryEntry.mkdir( );
else {
int count;
byte data[] = new byte[BUFFER];
// write the files to the disk
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( );
catch( IOException ioe ) {
System.out.println( "An IO exception has occurred:" );
ioe.printStackTrace( );
continue;
zis.close( );
catch( Exception e ) {
System.out.println( "A general exception occurred:" );
e.printStackTrace( );
}When I run the program, everything works fine until I get to the zip file entry causing the problem. The output looks like this:
An IO exception has occurred:
java.util.zip.ZipException: invalid entry size (expected 4007 but got 4055 bytes)
at java.util.zip.ZipInputStream.readEnd(ZipInputStream.java:367)
at java.util.zip.ZipInputStream.read(ZipInputStream.java:141)
at java.util.zip.ZipInputStream.closeEntry(ZipInputStream.java:91)
at java.util.zip.ZipInputStream.getNextEntry(ZipInputStream.java:69)
at ZipExtractor.main(ZipExtractor.java:23)
An IO exception has occurred:
java.io.IOException: Push back buffer is full
at java.io.PushbackInputStream.unread(PushbackInputStream.java:204)
at java.util.zip.ZipInputStream.readEnd(ZipInputStream.java:348)
at java.util.zip.ZipInputStream.read(ZipInputStream.java:141)
at java.util.zip.ZipInputStream.closeEntry(ZipInputStream.java:91)
at java.util.zip.ZipInputStream.getNextEntry(ZipInputStream.java:69)
at ZipExtractor.main(ZipExtractor.java:23)
An IO exception has occurred:
java.io.IOException: Push back buffer is full
at java.io.PushbackInputStream.unread(PushbackInputStream.java:204)
at java.util.zip.ZipInputStream.readEnd(ZipInputStream.java:348)
at java.util.zip.ZipInputStream.read(ZipInputStream.java:141)
at java.util.zip.ZipInputStream.closeEntry(ZipInputStream.java:91)
at java.util.zip.ZipInputStream.getNextEntry(ZipInputStream.java:69)
at ZipExtractor.main(ZipExtractor.java:23)The second error repeats itself until I ctrl-c out of the program.
What's happening is that the inner catch block detects the zip entry error as expected. It continues on the next pass through the while loop and attempts to read the next entry, whereupon the second IOException is thrown over and over again. (Incidentally, I've also tried using java.util.zip.ZipFile, but it won't even open the file.)
My questions are these:
1. What exactly does "Push back buffer is full" mean? I've looked through the javadocs, but my understanding is still sketchy. Can I manipulate the buffer or otherwise code around the problem in order to prevent this error from being thrown?
2. Is there any way I can see the source code for java.util.zip.ZipInputStream? It would be very helpful to see exactly where and why this error is being thrown.
Any help anyone can provide would be greatly appreciated. By the way, if this question should be better posted in another forum, let me know and I'll re-post it there.No clue. Too few info.
Post a small demo code that is generally compilable, runnable and could reproduce your problem. See: http://homepage1.nifty.com/algafield/sscce.html and http://www.yoda.arachsys.com/java/newsgroups.html -
Compressing a folder creates two zip files
I've got a folder called Blanks/ABC FM/SMS/Other and when I compress it I get two zip files of the same size: one with the correct name and the other called Archive.zip. When I remove the space I only get the Archive.zip. When I remove the slashes I get the correct zip file only.
Is this a bug that I should report? How?If you were expecting C/D to compress to C/D.zip instead of Archive.zip I think I have an explanation for that.
The unix file system which is the underlying OSX file system uses slash (/) as a path delimiter. So to use an example like yours, if you create a folder A on the desktop and B within A then that the pathname is A/B. But in reality, being on your desktop it is /Users/you/A/B, where you is your user id.
The finder tries to allow as many characters as it can for filenames. But the slash it cannot allow in a filename when recorded in the file system. So what it does is actually change the slash to a colon in the actual filename. Thus the filename "A/B" in the finder is, in reality, "A:B" in the file system.
I think this colon-equals-slash naming convention is what causing the finder's Compress to treat the name specially and it just changes it the output to be the generic Archive.zip.
Compress is there as a convenience. There's lots of other ways to do zips. There's third party utilities. There's zip command in the terminal (where you can play the finder's game and generate A:B.zip so that it displays in the finder as A/B.zip). -
Description of zip file for example B24897.zip
Hi,
Is below softwares 64 bit?
how can I get description of zip file for example B24897-01_3of4.zip is meant for what software?
10g Release 2 (10.2) for Microsoft Windows (x64)
file:///C:/oracle%2010g/B24971-01/welcome.html
Oracle Database
10g Release 2 (10.2) for Microsoft Windows (x64)
file:///C:/oracle%2010g/B24897-01_3of4/welcome.html
10g Release 2 (10.2) for Microsoft Windows (x64)
file:///C:/oracle%2010g/B24897-01_2of4/clusterware/welcome.html
10g Release 2 (10.2) for Microsoft Windows (x64)
file:///C:/oracle%2010g/B24897-01_4of4/database/welcome.html
Oracle Database
10g Release 2 (10.2) for Microsoft Windows (x64)
file:///C:/oracle%2010g/B24897-01_1of4/client/welcome.htmlHi;
Its:
B24897-01 Oracle Database 10g Release 2 (10.2.0.1.0) for Microsoft Windows x64 DVD
Please see:
http://pias.oracle.co.jp/pias_owa/pias/par_disp.product_comp?p_product_code=20394&v_current=2
Regard
Helios -
Installing JDK 1.4, esp. the src.zip file
I have some questions about installing JDK 1.4:
1. After running the executable installation file, you get a src.zip file in the installation directory. Does anyone know what this is? The obvious answer is that it is the library of code for JDK 1.4. However, I have run programs on Forte Agent without unzipping this file.
2. In previous versions of the JDK, when you decompress the src.jar (it is now a src.zip file in the 1.4 version) file, it creates a src folder and decompresses all the information into that. However, in JDK 1.4, it does not create a src folder, but rather, decompresses the files into the installation directory. Is this a fault of the new release or should I create a src folder and have the src.zip file decompress into that folder?
I am running Windows 2000.I have some questions about installing JDK 1.4:
1. After running the executable installation file, you
get a src.zip file in the installation directory. Does
anyone know what this is? The obvious answer is that
it is the library of code for JDK 1.4. However, I have
run programs on Forte Agent without unzipping this
file.The src.zip is the source code for the JDK, not library. Note: The file is not the whole source code, it does not have any native library so you cannot compile it.
>
2. In previous versions of the JDK, when you
decompress the src.jar (it is now a src.zip file in
the 1.4 version) file, it creates a src folder and
decompresses all the information into that. However,
in JDK 1.4, it does not create a src folder, but
rather, decompresses the files into the installation
directory. Is this a fault of the new release or
should I create a src folder and have the src.zip file
decompress into that folder?
I am running Windows 2000.No, it is just how Sun zip it. Anyway, as long as you don't care to look at the source, you can just leave it along. -
Zip: concatening compressed chunks, will it give a single zip file
Are concatenated zipped chunks recognized as a single zip file?
What I am trying to achieve(using java.util.zip) is like, my application is generating a large CSV file, i need to compress this data and write it into a file or send to network. But I dont want to wait until the complete data generation and need to compress data in chunks; assuming that if i append my compressed chunks to a file at last i will get a valid & complete zip file.
For testing this I made a sample program like:
//Here a.zip b.zip are files compressed using java.util.zip; here they act as
// data chunk (actually these will be genarted by my
// application dynamically & in parallel to this process)
FileInputStream f1 = new FileInputStream(new File("c://a.zip"));
FileInputStream f2 = new FileInputStream(new File("c://b.zip"));
// Opening file in append mode
FileOutputStream fos = new FileOutputStream("c://c.zip",true);
byte[] data = new byte[(int)new File("c://a.zip").length()];
f1.read(data);
fos.write(data);
byte[] data2 = new byte[(int)new File("c://b.zip").length()];
f2.read(data2);
fos.write(data2);
f1.close();
f2.close();
fos.close();
But this program does not work, as m not getting the final zip file containing the whole data, rather it gives data only from first file.
Can any one point out the error.
Or suggest me the better way to achieve this.hi
i want to sugest you like:--
say fileoutputstreem.flesh();
call this flesh() method . It is just like commit in DB.
Through this U may salve your problem.
You are creating zip files like a.zip and b.zip if you save in your c:\\ it will take some IO Operations. It is not sugested.
If you want to like that only i will sujest you to delete them after your process is completed. -
Getting a StackOverflowError trying to download big content sync zip file
We've noticed this happens when trying to download a big Content Sync zip file (typically of around 280MB, although recently we saw the same issue with a 250MB file, so we're not sure of the root cause of this).
The only prerequisite to reproduce this issue is to configure your content sync to have > 250MB of content (in the form of images, html, js, etc). The steps to reproduce are the following:
Navigate to the content sync console URL: /libs/cq/contentsync/content/console.html
Click on the 'Clear Cache' button.
Click on the 'Update Cache' button. No problems up to here, content sync cache (under /var/contentsync) is populated with expected assets and files.
Click on the 'Download Full' button. Almost immediately, the app crashes with the following stack trace:
02.05.2013 00:56:14.248 *ERROR* [204.90.11.3 [1367455869892] GET /etc/contentsync/audiusa-retail.zip HTTP/1.1] org.apache.sling.engine.impl.SlingRequestProcessorImpl service: Uncaught Throwable java.lang.StackOverflowError
at org.apache.commons.collections.map.AbstractReferenceMap.isEqualKey(AbstractReferenceMap.j ava:434)
at org.apache.commons.collections.map.AbstractHashedMap.getEntry(AbstractHashedMap.java:436)
at org.apache.commons.collections.map.AbstractReferenceMap.getEntry(AbstractReferenceMap.jav a:405)
at org.apache.commons.collections.map.AbstractReferenceMap.get(AbstractReferenceMap.java:230 )
at org.apache.jackrabbit.core.state.ItemStateReferenceCache.retrieve(ItemStateReferenceCache .java:147)
at org.apache.jackrabbit.core.state.LocalItemStateManager.getItemState(LocalItemStateManager .java:171)
at org.apache.jackrabbit.core.state.XAItemStateManager.getItemState(XAItemStateManager.java: 260)
at org.apache.jackrabbit.core.state.SessionItemStateManager.getItemState(SessionItemStateMan ager.java:161)
at org.apache.jackrabbit.core.ItemManager.getItemData(ItemManager.java:382)
at org.apache.jackrabbit.core.ItemManager.getItem(ItemManager.java:328)
at org.apache.jackrabbit.core.ItemManager.getItem(ItemManager.java:622)
at org.apache.jackrabbit.core.LazyItemIterator.prefetchNext(LazyItemIterator.java:122)
at org.apache.jackrabbit.core.LazyItemIterator.<init>(LazyItemIterator.java:104)
at org.apache.jackrabbit.core.LazyItemIterator.<init>(LazyItemIterator.java:85)
at org.apache.jackrabbit.core.ItemManager.getChildProperties(ItemManager.java:816)
at org.apache.jackrabbit.core.NodeImpl$10.perform(NodeImpl.java:2178)
at org.apache.jackrabbit.core.NodeImpl$10.perform(NodeImpl.java:2174)
at org.apache.jackrabbit.core.session.SessionState.perform(SessionState.java:216)
at org.apache.jackrabbit.core.ItemImpl.perform(ItemImpl.java:91)
at org.apache.jackrabbit.core.NodeImpl.getProperties(NodeImpl.java:2174)
at javax.jcr.util.TraversingItemVisitor.visit(TraversingItemVisitor.java:202)
at org.apache.jackrabbit.core.NodeImpl.accept(NodeImpl.java:1697)
at javax.jcr.util.TraversingItemVisitor.visit(TraversingItemVisitor.java:219)
at org.apache.jackrabbit.core.NodeImpl.accept(NodeImpl.java:1697)
at javax.jcr.util.TraversingItemVisitor.visit(TraversingItemVisitor.java:219)
at org.apache.jackrabbit.core.NodeImpl.accept(NodeImpl.java:1697)
at javax.jcr.util.TraversingItemVisitor.visit(TraversingItemVisitor.java:219)
at org.apache.jackrabbit.core.NodeImpl.accept(NodeImpl.java:1697)
at javax.jcr.util.TraversingItemVisitor.visit(TraversingItemVisitor.java:219)
We've extended the content sync framework by writing 2 ContentUpdateHandler implementations. Given that the content update process finishes correctly, we dont' think this is the root cause of the problem.
Any help on this subject is appreciated.
Thanks,
-DanielHi Daniel,
File daycare with steps to reproduce, logs, thread dump & hs_err file.
Thanks,
Sham
@adobe_sham -
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.
AndreI'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());
} -
Error in reading the contents of a zip file
EHello Experts,
I want to read the contents of a zip file.I have written the following program which reads the contents of a file named "index.xml" which resides in ReadZip.zip.My problem is , it is reading only the first line of that file & after that it is giving this error.
java.io.IOException: Stream closed
at java.util.zip.ZipInputStream.ensureOpen(ZipInputStream.java:43)
at java.util.zip.ZipInputStream.getNextEntry(ZipInputStream.java:67)
at components.ReadZipFile2.main(ReadZipFile2.java:26)
public class ReadZipFile2 {
public static void main(String args[]) {
try {
FileInputStream fis = new FileInputStream("C:\\ReadZip.zip");
ZipInputStream zis = new ZipInputStream(fis);
ZipEntry ze;
while ((ze = zis.getNextEntry()) != null) {
System.out.println(ze.getName());
if (ze.getName().equals("ReadZip/index.xml")) {
long size = ze.getSize();
if (size > 0) {
System.out.println("Length is " + size);
BufferedReader br = new BufferedReader(
new InputStreamReader(zis));
String line;
while ((line = br.readLine()) != null) {
System.out.println(line);
br.close();
} catch (IOException e) {
e.printStackTrace();
}It seems that zis is getting close after reading the first entry of file.I am unable to guess the reason for this.Please help.Thanx in advance.[redacted confused advice]
[_Compressing and Decompressing Data using Java - with many code samples_|http://java.sun.com/developer/technicalArticles/Programming/compression/|Yes Virginia, there really are code samples] -
After attaching zip file in cv01n its not getting saved
after attaching zip file in cv01n its not getting saved
Hi Jameel,
Please be clear, I understand that you are using Content Server right?
OK, then have you defined Content Repository?
Follow the Steps below:
1. Goto "OAC0" & Create your own Content Repository like say (Z_Jameel_CS)
give the proper description,
Document Area as "DMS Document Management System"
Storage Type - > 04HTTP Content Server
Protocol--> -
Version no --->0046/0047 ( whatever you are using)
HTTP Server -
> (Your content server ip address eg: 139.100.31.311)
PortNumber--->(ex 1088)
HTTP Script--> ContentServer/ContentServer.dll
2. Click on "CS Admin" Content Server Administration.
Ask your Basis guy to give you the User ID and Password for the Content Server
go to "Certificate" Tab,
you can see your Content Repository created Click on the "Mail" icon to get the Certificate.
then just check in the Settings & Statics tabs that all information has come or not.
3.
Then Define Content Category.TCode "OACT"
You need to create a New Category say Z_JAM
Document Area--> DMS
Content Repository---> (mention what you created earlier i.e. Z_Jameel_CS)
then click on the next icon what you see next this field.
everything from this Content Repostory will be Copied.
Then checkIn your documents in the Content category defined.
Regards
Rehman Khan
Reward Your Points First. -
Questions about the content of download meeting recording .zip file
I tried posting this on the resurrected Connect forum, but my Adobe ID wasn't recognized there....
Concerning the files that are included in the .zip file of the meeting recording that can be downloaded:
1) Is there any documentation describing the files and their contents (i.e. what each file represents, what each XML element and attribute in those files represent)
2) Are there any files that capture mouse movement on a shared desktop?
Thank you!Hi Sean,
Regarding your first post:
Thanks Jorma! I don't have access to an FMS build at the moment but I'm quite certain it's there. As for contacting Jaydeep, I am 90% sure he authorized us to broadcast his email on here if folks had questions about the tool, but, in the case that I'm wrong and he didn't - I'm going to double-check first.
Regarding your most recent post..
"To be clear, the most critical goal I'm trying to accomplish is to create an automated process that will download the recording meeting at its highest quality in a consistent and reliable manner".
I personally believe this is possible; unfortunately, I haven't seen it done yet. If your recording contains:
- audio
- a camera feed
- screensharing
Then I think you might be able to get this going. If it contains shared content, like a shared PPT, this gets trickier.
"To do this, of course, I have to reproduce some of the functionality that Connect provides, starting and combining video and audio streams according to the instructions in the control files."
Exactly right. If your recording didn't contain shared content, then all you've got on your hands are a bunch of audio/video files that you could edit together as you wanted with your favourite video editing tool. If it contains shared content, here's (at a high level) what's happening.
For shared PPTs or FTContent files:
First (for version 9 recordings only), Connect reads the information on the Shared Content's location and SCO within mainstream and indexstream and validates it before loading it. I don't recall this happening to the same extent with version 8 or earlier, but maybe it was. Now, if the content is validated (ie. Connect can find it) the share pod will display as black, if it doesn't, you get an empty pod with an message like "No content is being shared" or something like that.
Connect then looks at the actual FTContent file, and loads the content that is to be shared using the file path and sco ID listed in here. It's important to note that the SCO ID and file path in here will likely not be the same as the original file you uploaded to your room, it's a new SCO id (I believe SCOs of this type are called referenced scos) and new path.
Now...if I was going to build some sort of player which would play all these files in one screen to make a recording...I might not want to use Connect's code here. If you know the file path to the shared content (from FTContent), you could easily view it with the content URL (conveniently also in FtContent). I'm not a coder, but I'm envisioning something like Presenter's GUI where you've got the presentation's content in the main area, and a video file (if there is one) playing back on the side.
Anyways, food for thought if you want to try to go about this. Connect recordings are incredibly complex and they come with a big learning curve, but if you can make sense of them the knowledge is quite valuable. -
File global_awm10g.zip has corrupted contents
The file GLOBAL_AWM10g.zip located on download.oracle.com/otn/java/olap has several invalid files in it. Four of the .xml files have invalid checksums and the global_awm10g_handson.doc and global_tables.dmp generate an "invalid compressed data to inflate" error when attempting to unzip them. Please refresh the contents of this .zip file, if possible.
Thanks,
JimI also get a CRC error when unzipping these same files.
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