Our current implementation on our ASP.net website doesn't support Zip64 which it needs to, so we're moving from System.IO.Compression over to DotNetZip:
https://github.com/DinoChiesa/DotNetZip
This small archive:
https://github.com/DinoChiesa/DotNetZip/files/10184907/Zip.zip
Throws the error:
Spanned archives with more than 65534 segments are not supported at this time.
The code sample simply attempts to open the Zip file:
using (var data = new MemoryStream(fileBytes.ToArray()))
{
using (var archive = Ionic.Zip.ZipFile.Read(data))
{
....
}
I'm a little unsure what the issue is here, is there an easy workaround or is there a better altnerative library?
You need to understand what does mean to have spanned zip file. That means that a zip is split into more files.
The file you have linked appears not to be such file:
Archive: Zip.zip
There is no zipfile comment.
End-of-central-directory record:
-------------------------------
Zip archive file size: 646370 (000000000009DCE2h)
Actual end-cent-dir record offset: 646272 (000000000009DC80h)
Expected end-cent-dir record offset: 646272 (000000000009DC80h)
(based on the length of the central directory and its expected offset)
This zipfile constitutes the sole disk of a single-part archive; its
central directory contains 25 entries.
The central directory is 3521 (0000000000000DC1h) bytes long,
and its (expected) offset in bytes from the beginning of the zipfile
is 642751 (000000000009CEBFh).
...
I think the problem is how you try to read the file with fileBytes.ToArray(). The data variable should be a filename and not fileBytes.ToArray().
If you look at the provided example, on how to read zip file, from the git you can see that on line 53 you get ZipFile zip = ZipFile.Read(args[0], options), where args[0] is a zip filename.
Here is the complete example form the git:
/ ReadZip.cs
//
// ----------------------------------------------------------------------
// Copyright (c) 2006-2009 Microsoft Corporation. All rights reserved.
//
// This example is released under the Microsoft Public License .
// See the license.txt file accompanying this release for
// full details.
//
// ----------------------------------------------------------------------
//
// This simple example utility simply reads a zip archive and extracts
// all elements in it, to the specified target directory.
//
// compile with:
// csc /target:exe /r:Ionic.Zip.dll /out:ReadZip.exe ReadZip.cs
//
// Wed, 29 Mar 2006 14:36
//
using System;
using Ionic.Zip;
namespace Ionic.Zip.Examples
{
public class ReadZip
{
private static void Usage()
{
Console.WriteLine("usage:\n ReadZip2 <zipfile> <unpackdirectory>");
Environment.Exit(1);
}
public static void Main(String[] args)
{
if (args.Length != 2) Usage();
if (!System.IO.File.Exists(args[0]))
{
Console.WriteLine("That zip file does not exist!\n");
Usage();
}
try
{
// Specifying Console.Out here causes diagnostic msgs to be sent to the Console
// In a WinForms or WPF or Web app, you could specify nothing, or an alternate
// TextWriter to capture diagnostic messages.
var options = new ReadOptions { StatusMessageWriter = System.Console.Out };
using (ZipFile zip = ZipFile.Read(args[0], options))
{
// This call to ExtractAll() assumes:
// - none of the entries are password-protected.
// - want to extract all entries to current working directory
// - none of the files in the zip already exist in the directory;
// if they do, the method will throw.
zip.ExtractAll(args[1]);
}
}
catch (System.Exception ex1)
{
System.Console.Error.WriteLine("exception: " + ex1);
}
}
}
}
EDIT - the above posted zip file still generates the above error. I have checked the source to find out where is the culprit:
It thinks it is a spanned archive based on this read. It tries to read a 2 bytes (that are converted to 16-bit unsigned integer ) from the block starting at 2nd position. If the converted value is ==0xFFFF then it considers the file as spanning file with more than 65534 segments. Probably there is some bug in the packing of the zip file which makes the DotNetZip fail.
There is the SharpZipLib nuget package you can use as an alternative.
The code below runs successfully with input the file you posted
static void Main(string[] args)
{
Console.WriteLine("Hello, World!");
string filename = "../../../Zip.zip";
FileStream fs = File.OpenRead(filename);
ICSharpCode.SharpZipLib.Zip.ZipFile zf = new ICSharpCode.SharpZipLib.Zip.ZipFile(fs);
foreach (ICSharpCode.SharpZipLib.Zip.ZipEntry zipEntry in zf)
{
string entryFileName = zipEntry.Name;
Console.WriteLine("File: " + entryFileName);
}
}
Output
Hello, World!
File: scripts/c3runtime.js
File: data.json
File: style.css
File: scripts/offlineclient.js
File: images/shared-0-sheet1.png
File: images/tiledbackground-sheet0.png
File: images/shared-0-sheet0.png
File: appmanifest.json
File: scripts/main.js
File: workermain.js
File: scripts/dispatchworker.js
File: scripts/jobworker.js
File: scripts/supportcheck.js
File: icons/icon-16.png
File: icons/icon-32.png
File: icons/icon-128.png
File: icons/icon-256.png
File: icons/icon-64.png
File: icons/icon-512.png
File: icons/loading-logo.png
File: index.html
File: arcade.json
File: scripts/register-sw.js
File: sw.js
File: offline.json
P.S. I have worked with both libraries in the past and I have found that DotNetZip is easier to work with, but for this case only SharpZipLib works :).
I have a resources folder/package in the root of my project, I "don't" want to load a certain File. If I wanted to load a certain File, I would use class.getResourceAsStream and I would be fine!! What I actually want to do is to load a "Folder" within the resources folder, loop on the Files inside that Folder and get a Stream to each file and read in the content... Assume that the File names are not determined before runtime... What should I do? Is there a way to get a list of the files inside a Folder in your jar File?
Notice that the Jar file with the resources is the same jar file from which the code is being run...
Finally, I found the solution:
final String path = "sample/folder";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());
if(jarFile.isFile()) { // Run with JAR file
final JarFile jar = new JarFile(jarFile);
final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
while(entries.hasMoreElements()) {
final String name = entries.nextElement().getName();
if (name.startsWith(path + "/")) { //filter according to the path
System.out.println(name);
}
}
jar.close();
} else { // Run with IDE
final URL url = Launcher.class.getResource("/" + path);
if (url != null) {
try {
final File apps = new File(url.toURI());
for (File app : apps.listFiles()) {
System.out.println(app);
}
} catch (URISyntaxException ex) {
// never happens
}
}
}
The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.
Try the following.
Make the resource path "<PathRelativeToThisClassFile>/<ResourceDirectory>" E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:
URL url = MyClass.class.getResource("resources/");
if (url == null) {
// error - missing folder
} else {
File dir = new File(url.toURI());
for (File nextFile : dir.listFiles()) {
// Do something with nextFile
}
}
You can also use
URL url = MyClass.class.getResource("/com/abc/package/resources/");
The following code returns the wanted "folder" as Path regardless of if it is inside a jar or not.
private Path getFolderPath() throws URISyntaxException, IOException {
URI uri = getClass().getClassLoader().getResource("folder").toURI();
if ("jar".equals(uri.getScheme())) {
FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap(), null);
return fileSystem.getPath("path/to/folder/inside/jar");
} else {
return Paths.get(uri);
}
}
Requires java 7+.
I know this is many years ago . But just for other people come across this topic.
What you could do is to use getResourceAsStream() method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.
I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar... on both the IDE and on jar (release version).
I found java.nio.file.DirectoryStream to work the best to iterate over directory contents over both local filesystem and jar.
String fooFolder = "/foo/folder";
....
ClassLoader classLoader = foofClass.class.getClassLoader();
try {
uri = classLoader.getResource(fooFolder).toURI();
} catch (URISyntaxException e) {
throw new FooException(e.getMessage());
} catch (NullPointerException e){
throw new FooException(e.getMessage());
}
if(uri == null){
throw new FooException("something is wrong directory or files missing");
}
/** i want to know if i am inside the jar or working on the IDE*/
if(uri.getScheme().contains("jar")){
/** jar case */
try{
URL jar = FooClass.class.getProtectionDomain().getCodeSource().getLocation();
//jar.toString() begins with file:
//i want to trim it out...
Path jarFile = Paths.get(jar.toString().substring("file:".length()));
FileSystem fs = FileSystems.newFileSystem(jarFile, null);
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(fs.getPath(fooFolder));
for(Path p: directoryStream){
InputStream is = FooClass.class.getResourceAsStream(p.toString()) ;
performFooOverInputStream(is);
/** your logic here **/
}
}catch(IOException e) {
throw new FooException(e.getMessage());
}
}
else{
/** IDE case */
Path path = Paths.get(uri);
try {
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(path);
for(Path p : directoryStream){
InputStream is = new FileInputStream(p.toFile());
performFooOverInputStream(is);
}
} catch (IOException _e) {
throw new FooException(_e.getMessage());
}
}
Another solution, you can do it using ResourceLoader like this:
import org.springframework.core.io.Resource;
import org.apache.commons.io.FileUtils;
#Autowire
private ResourceLoader resourceLoader;
...
Resource resource = resourceLoader.getResource("classpath:/path/to/you/dir");
File file = resource.getFile();
Iterator<File> fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
load(fi.next())
}
If you are using Spring you can use org.springframework.core.io.support.PathMatchingResourcePatternResolver and deal with Resource objects rather than files. This works when running inside and outside of a Jar file.
PathMatchingResourcePatternResolver r = new PathMatchingResourcePatternResolver();
Resource[] resources = r.getResources("/myfolder/*");
Then you can access the data using getInputStream and the filename from getFilename.
Note that it will still fail if you try to use the getFile while running from a Jar.
As the other answers point out, once the resources are inside a jar file, things get really ugly. In our case, this solution:
https://stackoverflow.com/a/13227570/516188
works very well in the tests (since when the tests are run the code is not packed in a jar file), but doesn't work when the app actually runs normally. So what I've done is... I hardcode the list of the files in the app, but I have a test which reads the actual list from disk (can do it since that works in tests) and fails if the actual list doesn't match with the list the app returns.
That way I have simple code in my app (no tricks), and I'm sure I didn't forget to add a new entry in the list thanks to the test.
Below code gets .yaml files from a custom resource directory.
ClassLoader classLoader = this.getClass().getClassLoader();
URI uri = classLoader.getResource(directoryPath).toURI();
if("jar".equalsIgnoreCase(uri.getScheme())){
Pattern pattern = Pattern.compile("^.+" +"/classes/" + directoryPath + "/.+.yaml$");
log.debug("pattern {} ", pattern.pattern());
ApplicationHome home = new ApplicationHome(SomeApplication.class);
JarFile file = new JarFile(home.getSource());
Enumeration<JarEntry> jarEntries = file.entries() ;
while(jarEntries.hasMoreElements()){
JarEntry entry = jarEntries.nextElement();
Matcher matcher = pattern.matcher(entry.getName());
if(matcher.find()){
InputStream in =
file.getInputStream(entry);
//work on the stream
}
}
}else{
//When Spring boot application executed through Non-Jar strategy like through IDE or as a War.
String path = uri.getPath();
File[] files = new File(path).listFiles();
for(File file: files){
if(file != null){
try {
InputStream is = new FileInputStream(file);
//work on stream
} catch (Exception e) {
log.error("Exception while parsing file yaml file {} : {} " , file.getAbsolutePath(), e.getMessage());
}
}else{
log.warn("File Object is null while parsing yaml file");
}
}
}
Took me 2-3 days to get this working, in order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
Whatever it is, in order for your JAR file to find it, the url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.
Simple ... use OSGi. In OSGi you can iterate over your Bundle's entries with findEntries and findPaths.
Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:
URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);
URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);
URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);
In downloading files using the following codes, it surprised me that it succeeded in downloading .jpg .txt files BUT .dgn format file return IO Error #2038
Could somebody give me any advice? Thanks in advance.
protected function init(event:FlexEvent):void
{
fileRef = new FileReference();
fileRef.addEventListener(Event.COMPLETE, doEvent);
fileRef.addEventListener(HTTPStatusEvent.HTTP_STATUS, doEvent);
fileRef.addEventListener(IOErrorEvent.IO_ERROR, doEvent);
fileRef.addEventListener(SecurityErrorEvent.SECURITY_ERROR, doEvent);
}
private function doEvent(evt:Event):void
{
var fr:FileReference = evt.currentTarget as FileReference;
switch (evt.type)
{
case "complete":
Alert.show("File : " + fr.name + " download succeed");
break;
default :
Alert.show("Error occur during downloading !!!");
break;
}
protected function downLoadLICMap(event:MouseEvent):void
{
urlReq = new URLRequest("http://svygis/viewphoto/ceddphoto/20130916 raw/1se19d.dgn");
fileRef.download(urlReq);
}
I suspect Flex is not at fault here. It would rather be a server setting issue.
Have you tried opening that URL directly in the browser? You probably will not be able to download the file like that either.
If that's the case, you only need to configure a mime type for the .dgn extension on your web server.
Here's the correct translation for this question, which was originally asked in french. Note that I have taken liberty to translate the comments in the code.
My procedure worked correctly during tests in my sandbox. When I put it into operational mode, it still worked correctly. Then I added a check in my code to prevent access to the sandbox version by copy/pasting the URL. Now my website works correctly in operational mode, but my sandbox vresion doesn't work anymore.
Link to operational mode.
Link to sandbox mode.
The procedure:
//
// VENDOR PARAMETERS FOR SANDBOX VERSION
//
if ($proctest == "1")
{
$url_nvp = 'https://api-3t.sandbox.paypal.com/nvp'; // Sandbox version
$version = 64.0; // Version
$iduser = 'f-facilitator_api1.x.fr'; // User
$passwrd = '0123456789'; // Password
// Signature
$signature = 'AFcWxV21C7fd0v3bYYYRCpSSRl31ALWKEzeddmFHrClYoc6tJpZiawjH';
}
//
// VENDOR PARAMETERS FOR OPERATIONAL VERSION
//
else
{
$url_nvp = 'https://api-3t.paypal.com/nvp'; // Operational website
$version = 64.0; // Version
$iduser = 'f_api1.x.fr'; // User
$passwrd = '0123456789'; // Password
// Signature
$signature = 'Apekq0Tf.isqMqkIsEX7RsjIFTVCA8EehX5M263oELbE40NBWWYxhtW1';
}
//
// BUILDING THE STRING
//
$api_paypal= $url_nvp.'?VERSION=' .$version // Builds the URL
.'&USER=' .$iduser
.'&PWD=' .$passwrd
.'&SIGNATURE=' .$signature;
return $api_paypal; // Returns the string
}
Then:
//
// TEST ENVIRONMENT FOR NVP'S API
//
if ($proctest == "1")
{
header("Location: https://www.sandbox.paypal.com/webscr&cmd=_express-checkout&token=".$liste_param_paypal['TOKEN']);
}
//
// OPERATIONAL ENVIRONMENT FOR NVP'S API
//
else
{
header("Location: https://www.paypal.com/webscr&cmd=_express-checkout&token=".$liste_param_paypal['TOKEN']);
}
I can't find what prevents the sandbox version from running correctly.
I recently upgraded to Flash Builder 4.5 for PHP and am trying to upload a release build to my remoteserver. When I try to make a php call from the app, I get the error:
Send failednChannel.Security.Error error Error #2048 url: 'http://localhost/my_php/public/gateway.php'
The release build works fine on my localhost machine. All of my php service calls are on my remote host. Here's the structure of my remote host:
/my_directory/html (this is the root directory)
/my_directory/html/my_php/public/release (this is where my .html wrapper and .swf files sit)
/my_directory/html/my_php/public (this is where my gateway.php and amf_config.ini files sit)
The error specifically references 'localhost', but I can't find where that gets set. When I google error #2048, the solutions point to a badly configured crossdomain file...all my services are on remotehost (where the app is hosted) so I don't think that could be the issue.
Here is my amf_config.ini file:
[zend]
webroot = "/my_directory/html"
zend_path ="/my_directory/html/ZendFramework/library"
library ="/my_directory/html/my_php/library"
services ="/my_directory/html/my_php/services"
[zendamf]
amf.production = false
amf.directories[]=/my_directory/html/my_php/services
Here is my gateway.php file:
<?php
ini_set("display_errors", 1);
$dir = dirname(__FILE__);
$webroot = $_SERVER['DOCUMENT_ROOT'];
$configfile = "$dir/amf_config.ini";
$servicesdir = $dir.'/../services';
$librarydir = $dir.'/../library';
//default zend install directory
$zenddir = $webroot.'/ZendFramework/library';
//Load ini file and locate zend directory
if (file_exists($configfile)) {
$arr = parse_ini_file($configfile, true);
if (isset($arr['zend']['webroot'])) {
$webroot = $arr['zend']['webroot'];
$zenddir = $webroot.'/ZendFramework/library';
}
if (isset($arr['zend']['zend_path'])) {
$zenddir = $arr['zend']['zend_path'];
}
if (isset($arr['zend']['library'])) {
$librarydir = $arr['zend']['library'];
}
if (isset($arr['zend']['services'])) {
$servicesdir = $arr['zend']['services'];
}
}
// Setup include path
// add zend directory, library and services to include path
set_include_path(get_include_path()
.PATH_SEPARATOR.$zenddir
.PATH_SEPARATOR.$librarydir
.PATH_SEPARATOR.$servicesdir);
// Initialize Zend Framework loader
require_once 'Zend/Loader/Autoloader.php';
Zend_Loader_Autoloader::getInstance()->setFallbackAutoloader(true)->suppressNotFoundWarnings(true);
// Load configuration
$default_config = new Zend_Config(array("production" => false), true);
$default_config->merge(new Zend_Config_Ini($configfile, 'zendamf'));
$default_config->setReadOnly();
$amf = $default_config->amf;
// Store configuration in the registry
Zend_Registry::set("amf-config", $amf);
// Initialize AMF Server
$server = new Zend_Amf_Server();
$server->setProduction($amf->production);
if (isset($amf->directories)) {
$dirs = $amf->directories->toArray();
foreach ($dirs as $dir) {
if ($dir == "./") {
$server->addDirectory($webroot);
} else
if (realpath("{$webroot}/{$dir}")) {
$server->addDirectory("{$webroot}/{$dir}");
} else
if (realpath($dir)) {
$server->addDirectory(realpath($dir));
}
}
}
// Initialize introspector for non-production
if (! $amf->production) {
$server->setClass('Zend_Amf_Adobe_Introspector', '',
array("config" => $default_config, "server" => $server));
$server->setClass('Zend_Amf_Adobe_DbInspector', '',
array("config" => $default_config, "server" => $server));
}
// Handle request
echo $server->handle();
i had the same problem in a flex - blaze - environment. The real Problem was the context-root in the project properties.
because you used flex 4.5, there is no input field for this setting. in flex builder 3, there was a setting in the project properties -> flex server -> context root.
i gone crazy and found after hours an article on adobes bugs-site [FB-22939].
That solve my problem. I have no idea, wich project settings you are using, try to search a string named {context.root} in your project or post a liite bit more about your project settings. i know blaze is different from php, but maybe it is a tip bringing you back on track.
Unfortunately I'm not able to reproduce my thread and setting up an php environment with more knowledge about your setup. (Server-Technology, and so one)
edit:
additional Info:
I've found a list of all compiler arguments. Try it with this argument:
-context-root <context-path>
full name -compiler.context-root
path to replace {context.root} tokens for service channel endpoints
br
Frank