Develop Reference

How to have my java project to use some files without using their absolute path?

I have written a project where some images are used for the application's appearance and some text files will get created and deleted along the process. I only used the absolute path of all used files in order to see how the project would work, and now that it is finished I want to send it to someone else. so what I'm asking for is that how I can link those files to the project so that the other person doesn't have to set those absolute paths relative to their computer. something like, turning the final jar file with necessary files into a zip file and then that the person extracts the zip file and imports jar file, when runs it, the program work without any problems.
by the way, I add the images using ImageIcon class.
I'm using eclipse.

For files that you just want to read, such as images used in your app's icons:
Ship them the same way you ship your class files: In your jar or jmod file.
Use YourClassName.class.getResource or .getResourceAsStream to read these. They are not files, any APIs that need a File object can't work. Don't use those APIs (they are bad) - good APIs take a URI, URL, or InputStream, which works fine with this.
Example:
package com.foo;
public class MyMainApp {
public void example() {
Image image = new Image(MyMainApp.class.getResource("img/send.png");
}
public void example2() throws IOException {
try (var raw = MyMainApp.class.getResourceAsStream("/data/countries.txt")) {
BufferedReader in = new BufferedReader(
new InputStreamReader(raw, StandardCharsets.UTF_8));
for (String line = in.readLine(); line != null; line = in.readLine()) {
// do something with each country
}
}
}
}
This class file will end up in your jar as /com/foo/MyMainApp.class. That same jar file should also contain /com/foo/img/send.png and /data/countries.txt. (Note how starting the string argument you pass to getResource(AsStream) can start with a slash or not, which controls whether it's relative to the location of the class or to the root of the jar. Your choice as to what you find nicer).
For files that your app will create / update:
This shouldn't be anywhere near where your jar file is. That's late 80s/silly windows thinking. Applications are (or should be!) in places that you that that app cannot write to. In general the installation directory of an application is a read-only affair, and most certainly should not be containing a user's documents. These should be in the 'user home' or possibly in e.g. `My Documents'.
Example:
public void save() throws IOException {
Path p = Paths.get(System.getProperty("user.home"), "navids-app.save");
// save to that file.
}

Get resources from a jar file

i want my jar file to access some files from itself. I know how to do this for BufferedImage but this doesn't work for other files. All i want is to extract some zips from my jar. i made a class folder in eclipse, put the zips inside and used
public File getResFile(String name){
return new File(getClass().getResource(name).getFile());
}
to get the File instance and extract it. it works fine in eclipse, but as soon as i export it to a jar it says
Exception in thread "main" java.io.FileNotFoundException: file:\C:\Users\DeLL\Desktop\BoxcraftClient\ClientInstaller.jar!\client.bxc (The filename, directory name, or volume label syntax is incorrect)
at java.util.zip.ZipFile.open(Native Method)
at java.util.zip.ZipFile.<init>(ZipFile.java:220)
at java.util.zip.ZipFile.<init>(ZipFile.java:150)
at java.util.zip.ZipFile.<init>(ZipFile.java:164)
at Launcher.install(Launcher.java:43)
at Launcher.main(Launcher.java:33)
Im working to fix this already something like 6 hours and can't find a solution. Please help!

There is a reason why getResource() returns a URL, and not a File, because the resource may not be a file, and since your code is packaged in the Jar file, it's not a file but a zip entry.
The only safe way to read the content of the resource, is as an InputStream, either by calling getResourceAsStream() or by calling openStream() on the returned URL.

first check your class path using java System.out.println("classpath is: " + System.getProperty("java.class.path")); to see if the classpath has your jar file.
And then use the getclass().classloader.getResourceAsStream(name). See if the returned URL is correct. Call the method isFile() on the URL to check if the URL is actually a file. And then call the getFile() method.

Use one of these methods, from the class Class
- getResource(java.lang.String)
- getResourceAsStream(java.lang.String)
this.getClass().getResource(name);
this.getClass().getResourceAsStream(name);
Warning: By default it loads the file from the location, where the this.class, is found in the package. So if using it from a class org.organisation.project.App, then the file need to be inside the jar in the directory org/organisation/project. In case the file is located in the root, or some other directory, inside the jar, use the /, in from of the file name. Like /data/names.json.

Use Spring's PathMatchingResourcePatternResolver;
It will do the trick for both launching the package from an IDE or from the file system:
public List<String> getAllClassesInRunningJar() throws Exception {
try {
List<String> list = new ArrayList<String>();
// Get all the classes inside the package com.my.package:
// This will do the work for both launching the package from an IDE or from the file system:
String scannedPackage = "com.my.package.*";
// This is spring - org.springframework.core; use these imports:
// import org.springframework.core.io.Resource;
// import org.springframework.core.io.support.PathMatchingResourcePatternResolver;
PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
Resource[] resources = scanner.getResources(scannedPackage.replace(".", "/"));
for (Resource resource : resources)
list.add(resource.getURI().toString());
return list ;
} catch (Exception e) {
throw new Exception("Failed to get the classes: " + e.getMessage(), e);
}
}

Property file not reflecting the modified changes using Apache Commons Configuration

I am trying to explore Apache commons configuration to dynamically load the property file and do modification in the file and save it.
I wrote a demo code for the same.
Code Snippet
package ABC;
import org.apache.commons.configuration.ConfigurationException;
import org.apache.commons.configuration.PropertiesConfiguration;
import org.apache.commons.configuration.reloading.FileChangedReloadingStrategy;
public class Prop {
public static void main(String[] args)
{
try {
URL propertiesURL = Prop.class.getResource("/d1.properties");
if (propertiesURL == null) {
System.out.println("null");
}
String absolutePath=propertiesURL.getPath();
PropertiesConfiguration pc = new PropertiesConfiguration(absolutePath);
pc.setReloadingStrategy(new FileChangedReloadingStrategy());
String s=(String)pc.getProperty("key_account_sales");
System.out.println("s is " + s);
pc.setAutoSave(true);
pc.setProperty("key_account_sales", "Dummy");
pc.save();
System.out.println("Modified as well");
String sa=(String)pc.getProperty("key_account_sales");
System.out.println("s is " + sa);
}catch(ConfigurationException ce)
{
ce.printStackTrace();
}
}
}
Although when I run the code multiple times, the updated value for the property is being properly shown but the changes are not seen in the Property file.
I tried refreshing the entire workspace and the project but still the property file shows the previous entry whereas this code displays the updated entry in console.
Why my property file is not getting updated?
Well I noticed that a new file with same name was formed inside bin
directory of my IDE workspace. This new file contains the required
changes.
However I still want that the old file should be updated with the new
value and instead of creating a new file, it should update in the old
file itself.
My property file is located inside a Web Application package say
Dem1
by the name of
Prop1.prop
I want to read this property file from in another class say
Reading.java
located inside another package
Dem2
, do changes in this same property file and show it to another user. It is a web application being deployed on an application server.
Even after using the absolute path in a simple file (main function) it is not reflecting the changes in the same file but updating it in new file.
I am doing a very slight mistake but can someone please help.
Using absolute path I am not able to make changes in the same property file in normal main method also. Please suggest.
New file in bin directory is created instead of updating the same file
in src folder.

You should be able to solve this using absolute paths. The PropertiesConfiguration class is finding your properties file somewhere on the classpath and only knows to write back to "d1.properties"; hence you have a file appearing in your bin directory.
The absolute path can be obtained by querying resources on the classpath. Something like the following:
URL propertiesURL = Prop.class.getResource("/d1.properties");
if (propertiesURL == null) {
// uh-oh...
}
String absolutePath = propertiesURL.getPath();
// Now use absolutePath

Getting a directory inside a .jar

I am trying to access a directory inside my jar file.
I want to go through every of the files inside the directory itself. I tried, for example, using the following:
URL imagesDirectoryURL=getClass().getClassLoader().getResource("Images");
if(imagesFolderURL!=null)
{
File imagesDirectory= new File(imagesDirectoryURL.getFile());
}
If I test this applet, it works well. But once I put the contents into the jar, it doesn't because of several reasons.
If I use this code, the URL always points outside the jar, so I have to put the Images directory there.
But if I use new File(imagesDirectoryURL.toURI());, it doesn't work inside the jar because I get the error URI not hierarchical. I am sure the directory exists inside the jar.
How am I supposed the get the contents of Images inside the jar?

Here is a solution which should work given that you use Java 7... The "trick" is to use the new file API. Oracle JDK provides a FileSystem implementation which can be used to peek into/modify ZIP files, and that include jars!
Preliminary: grab System.getProperty("java.class.path", "."), split against :; this will give you all entries in your defined classpath.
First, define a method to obtain a FileSystem out of a classpath entry:
private static final Map<String, ?> ENV = Collections.emptyMap();
//
private static FileSystem getFileSystem(final String entryName)
throws IOException
{
final String uri = entryName.endsWith(".jar") || entryName.endsWith(".zip"))
? "jar:file:" + entryName : "file:" + entryName;
return FileSystems.newFileSystem(URI.create(uri), ENV);
}
Then create a method to tell whether a path exists within a filesystem:
private static boolean pathExists(final FileSystem fs, final String needle)
{
final Path path = fs.getPath(needle);
return Files.exists(path);
}
Use it to locate your directory.
Once you have the correct FileSystem, use it to walk your directory using .getPath() as above and open a DirectoryStream using Files.newDirectoryStream().
And don't forget to .close() a FileSystem once you're done with it!
Here is a sample main() demonstrating how to read all the root entries of a jar:
public static void main(final String... args)
throws IOException
{
final Map<String, ?> env = Collections.emptyMap();
final String jarName = "/opt/sunjdk/1.6/current/jre/lib/plugin.jar";
final URI uri = URI.create("jar:file:" + jarName);
final FileSystem fs = FileSystems.newFileSystem(uri, env);
final Path dir = fs.getPath("/");
for (Path entry : Files.newDirectoryStream(dir))
System.out.println(entry);
}

Paths within Jars are paths, not actual directories as you can use them on a file system. To get all resources within a particular path of a Jar file:
Gain an URL pointing to the Jar.
Get an InputStream from the URL.
Construct a ZipInputStream from the InputStream.
Iterate each ZipEntry, looking for matches to the desired path.
..will I still be able to test my Applet when it's not inside that jar? Or will I have to program two ways to get my Images?
The ZipInputStream will not work with loose resources in directories on the file system. But then, I would strongly recommend using a build tool such as Ant to build (compile/jar/sign etc.) the applet. It might take an hour or so to write the build script & check it, but thereafter you can build the project by a few keystrokes and a couple of seconds.
It would be quite annoying if I always have to extract and sign my jar if I want to test my Aplet
I'm not sure what you mean there. Where does the 'extract' come into it? In case I was not clear, a sand-boxed applet can load resources this way, from any Jar that is mentioned in the archive attribute. Another thing you might do, is to separate the resource Jar(s) from the applet Jar. Resources typically change less than code, so your build might be able to take some shortcuts.
I think I really have to consider putting my Images into a seperate directory outside the jar.
If you mean on the server, there will be no practical way to get a listing of the image files short of help from the server. E.G. Some servers are insecurely set up to produce an HTML based 'file list' for any directory with no default file (such as an index.html).
I have only got one jar, in which my classes, images and sounds are.
OK - consider moving the sounds & images into a separate Jar. Or at the very least, put them in the Jar with 'no compression'. While Zip comression techniques work well with classes, they are less efficient at compressing (otherwise already compressed) media formats.
I have to sign it because I use the "Preferences" class to save user settings."
There are alternatives to the Preferences for applets, such as cookies. In the case of plug-in 2 architecture applet, you can launch the applet (still embedded in the browser) using Java Web Start. JWS offers the PersistenceService. Here is my small demo. of the PersistenceService.
Speaking of JWS, that brings me to: Are you absolutely certain this game would be better as an applet, rather than an app (e.g. using a JFrame) launched using JWS?
Applets will give you no end of stress, and JWS has offered the PersistenceService since it was introduced in Java 1.2.

You can use the PathMatchingResourcePatternResolver provided by Spring.
public class SpringResourceLoader {
public static void main(String[] args) throws IOException {
PathMatchingResourcePatternResolver resolver = new PathMatchingResourcePatternResolver();
// Ant-style path matching
Resource[] resources = resolver.getResources("/Images/**");
for (Resource resource : resources) {
System.out.println("resource = " + resource);
InputStream is = resource.getInputStream();
BufferedImage img = ImageIO.read(is);
System.out.println("img.getHeight() = " + img.getHeight());
System.out.println("img.getWidth() = " + img.getWidth());
}
}
}
I didn't do anything fancy with the returned Resource but you get the picture.
Add this to your maven dependency (if using maven):
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-core</artifactId>
<version>3.1.2.RELEASE</version>
</dependency>
This will work directly from within Eclipse/NetBeans/IntelliJ and in the jar that's deployed.
Running from within IntelliJ gives me the following output:
resource = file [C:\Users\maba\Development\stackoverflow\Q12016222\target\classes\pictures\BMW-R1100S-2004-03.jpg]
img.getHeight() = 768
img.getWidth() = 1024
Running from command line with executable jar gives me the following output:
C:\Users\maba\Development\stackoverflow\Q12016222\target>java -jar Q12016222-1.0-SNAPSHOT.jar
resource = class path resource [pictures/BMW-R1100S-2004-03.jpg]
img.getHeight() = 768
img.getWidth() = 1024

I think you can directly access resources in ZIP/JAR file
Please see Tutorial its giving solution to your question
How to extract Java resources from JAR and zip archives
Hopes that helps

If I understand your problem you want to check the directory inside the jar and check all the files inside that directory.You can do something like:
JarInputStream jar = new JarInputStream(new FileInputStream("D:\\x.jar"));
JarEntry jarEntry ;
while(true)
{
jarEntry = jar.getNextJarEntry();
if(jarEntry != null)
{
if(jarEntry.isDirectory() == false)
{
String str = jarEntry.getName();
if(str.startsWith("weblogic/xml/saaj"))
{
anything which comes here are inside weblogic\xml\saaj directory
}
}
}
}

What you are looking for here might be the JarEntry list of the Jar... I had done some similar work during grad school... You can get the modified class here (http://code.google.com/p/marcellodesales-cs-research/source/browse/trunk/grad-ste-ufpe-brazil/ptf-add-on-dev/src/br/ufpe/cin/stp/global/filemanager/JarFileContentsLoader.java) Note that the URL contains an older Java class not using Generics...
This class returns a set of URLs with the protocol "jar:file:/" for a given token...
package com.collabnet.svnedge.discovery.client.browser.util;
import java.io.IOException;
import java.net.URL;
import java.util.Enumeration;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;
import java.util.jar.JarEntry;
import java.util.jar.JarFile;
public class JarFileContentsLoader {
private JarFile jarFile;
public JarFileContentsLoader(String jarFilePath) throws IOException {
this.jarFile = new JarFile(jarFilePath);
}
/**
* #param existingPath an existing path string inside the jar.
* #return the set of URL's from inside the Jar (whose protocol is "jar:file:/"
*/
public Set<URL> getDirEntries(String existingPath) {
Set<URL> set = new HashSet<URL>();
Enumeration<JarEntry> entries = jarFile.entries();
while (entries.hasMoreElements()) {
String element = entries.nextElement().getName();
URL url = getClass().getClassLoader().getResource(element);
if (url.toString().contains("jar:file")
&& !element.contains(".class")
&& element.contains(existingPath)) {
set.add(url);
}
}
return set;
}
public static void main(String[] args) throws IOException {
JarFileContentsLoader jarFileContents = new JarFileContentsLoader(
"/u1/svnedge-discovery/client-browser/lib/jmdns.jar");
Set<URL> entries = jarFileContents.getDirEntries("impl");
Iterator<URL> a = entries.iterator();
while (a.hasNext()) {
URL element = a.next();
System.out.println(element);
}
}
}
The output would be:
jar:file:/u1/svnedge-discovery/client-browser/lib/jmdns.jar!/javax/jmdns/impl/constants/
jar:file:/u1/svnedge-discovery/client-browser/lib/jmdns.jar!/javax/jmdns/impl/tasks/state/
jar:file:/u1/svnedge-discovery/client-browser/lib/jmdns.jar!/javax/jmdns/impl/tasks/resolver/
jar:file:/u1/svnedge-discovery/client-browser/lib/jmdns.jar!/javax/jmdns/impl/
jar:file:/u1/svnedge-discovery/client-browser/lib/jmdns.jar!/javax/jmdns/impl/tasks/

May the following code sample can help you
Enumeration<URL> inputStream = BrowserFactory.class.getClassLoader().getResources(".");
System.out.println("INPUT STREAM ==> "+inputStream);
System.out.println(inputStream.hasMoreElements());
while (inputStream.hasMoreElements()) {
URL url = (URL) inputStream.nextElement();
System.out.println(url.getFile());
}

IF you really want to treat JAR files like directories, then please have a look at TrueZIP 7. Something like the following might be what you want:
URL url = ... // whatever
URI uri = url.toURI();
TFile file = new TFile(uri); // File-look-alike in TrueZIP 7
if (file.isDirectory) // true for regular directories AND JARs if the module truezip-driver-file is on the class path
for (TFile entry : file.listFiles()) // iterate top level directory
System.out.println(entry.getPath()); // or whatever
Regards,
Christian

How can I get relative path of the folders in my android project?

How can I get the relative path of the folders in my project using code?
I've created a new folder in my project and I want its relative path so no matter where the app is, the path will be correct.
I'm trying to do it in my class which extends android.app.Activity.
Perhaps something similar to "get file path from asset".

Make use of the classpath.
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
URL url = classLoader.getResource("path/to/folder");
File file = new File(url.toURI());
// ...

Are you looking for the root folder of the application? Then I would use
String path = getClass().getClassLoader().getResource(".").getPath();
to actually "find out where I am".

File relativeFile = new File(getClass().getResource("/icons/forIcon.png").toURI());
myJFrame.setIconImage(tk.getImage(relativeFile.getAbsolutePath()));

With this I found my project path:
new File("").getAbsolutePath();
this return "c:\Projects\SampleProject"

You can check this sample code to understand how you can access the relative path using the java sample code
import java.io.File;
public class MainClass {
public static void main(String[] args) {
File relative = new File("html/javafaq/index.html");
System.out.println("relative: ");
System.out.println(relative.getName());
System.out.println(relative.getPath());
}
}
Here getPath will display the relative path of the file.

In Android, application-level meta data is accessed through the Context reference, which an activity is a descendant of.
For example, you can get the source directory via the getApplicationInfo().sourceDir property.
There are methods for other folders as well (assets directory, data dir, database dir, etc.).

Generally we want to add images, txt, doc and etc files inside our Java project and specific folder such as /images.
I found in search that in JAVA, we can get path from Root to folder which we specify as,
String myStorageFolder= "/images"; // this is folder name in where I want to store files.
String getImageFolderPath= request.getServletContext().getRealPath(myStorageFolder);
Here, request is object of HttpServletRequest. It will get the whole path from Root to /images folder. You will get output like,
C:\Users\STARK\Workspaces\MyEclipse.metadata.me_tcat7\webapps\JavaProject\images

With System.getProperty("user.dir") you get the "Base of non-absolute paths" look at
Java Library Description