Develop Reference

How do you load an Image from a ZIP file contained in a .jar file?

We are programming a game, which shall be startable from a .jar file. First we created a Project in IntelliJ and loaded the Images from a ZIP with the following code:
ZipFile zf = null;
try {
zf = new ZipFile(zipPath);
Image Image = ImageIO.read(zf.getInputStream(zf.getEntry("Block/Air.png")));
} catch (IOException ignored) {}
Now the attempt without the ZIP (just from the .jar) is:
Image image=ImageIO.read(getClass().getResourceAsStream(path+ "Block/Air.png"));
It doesn't load any texture. Do you have a better way to do this in combination?
Edit:Seems not to be the Problem.

Since jars are zip files you could place them in the jar file and placing them the classpath.
Image image=ImageIO.read(getClass().getResourceAsStream(path+ "Block/Air.png");
Path must be a relative path from a source path root. E.g. I have a file in "src/main/resource/my/cool/game/" path is "/my/cool/game".
If you want to use a zip file, it must be outside of your jar file. To load the zip, you could use a relative file path, which is the same if you start your game from Intellij and from dekstop.
To change the working directory in intellij look here.
The best way would be to place the zip file alongside the jar so you can use "." as working directory to load the zip file.
Alternatively you could use a fixed directory, but then your game needs some sort of installation so it knows where to finde the zip file.
If you use ".", the zip file needs to be in the root of the project directory.

The Class.getResource and Class.getResourceAsStream methods take a URL (not a file path!) which is relative to the root of each classpath entry. For classpath entries which are .jar files, this means the path of a file packaged within the respective .jar file.
If your entire program is in one .jar file, the classpath consists of just one item: that .jar file. Therefore, there is only one classpath root, and the String you pass to getResourceAsStream is the URL of an entry within your .jar file. Do not include the path to the .jar file in that String.
If you are not sure what you should pass, examine your .jar file's contents. Every IDE (that I know of) provides a way to do this. You can also use any unzip utility to examine a .jar file, since every .jar file is actually a .zip file. (If you only have Windows, with no zip tools installed, make a copy of the .jar file and change the copy's extension to ".zip", then open it.)
Inside the .jar file are, of course, zip entries. The full path of the entry you want to load (without the path to the .jar file) is what you must pass to getResourceAsStream. getResourceAsStream accepts a URL, and URLs always use forward slashes (/) on all platforms, so do not use any backslashes. Also, the first character of the String must be /.
It is actually possible to specify a shorter path, depending on how your images are packaged in the .jar, but that is a separate topic. See the documentation for full details.
Side note: Never, ever write an empty catch block. Ever. That caught exception is by far the easiest way for you or anyone else to know when and why your program is not working. At the very least, put exc.printStackTrace(); in your catch block. More often, the correct course of action is to abort the program with something like throw new RuntimeException(exc);. After all, your program can't continue to function properly if it can't load that image, right?

Why do you need to store your images in a zip file? If you're doing it to reduce the file size, you gain absolutely nothing from zipping it first. JAR files are zipped files anyway (if you don't believe me, rename your .jar file to .zip, and try to open it). What you're basically doing is attempting to zip an already zipped file, which doesn't really do anything.
I would recommend you unzip your images and store them somewhere like < resources >/images
If you insist on leaving them zipped, you'll need to change it to something like this. Otherwise, it's looking for the zip file in the working directory (directory from which the jar was executed)
ZipFile zf = new ZipFile(getClass().getResourcesAsStream("path/to/zip"));
Disclaimer: I am not familiar with the ZipFile class, so I do not know if that constructor exists.

How to make class files into a JAR file

I have a JAR file that I wanted to edit a string inside on of it's classes . So I extracted it using Winrar , done the changes I wanted and saved it . Now I have a folder with subfolders inside that contains class files (about 30 one) .My question here is how can I recreate the Jar file from those folders&files ? Like it's it's doing the reverse action of extracting the Jar file .. ?

You could create a .zip file, rename it (with WinRAR) to something.jar so it becomes a .jar archive and then copy the folders in there.
But what you are doing is not really recommended to be done
You shouldn't extract and re-archive it like that.
And DON'T I repeat DON'T edit .class files!

Jar file is basically a zip file - however the best way to edit jars is to use a file manager (like Total Commander). You can delete, copy, rename, ... whatever, transparently inside the archive.
http://en.wikipedia.org/wiki/JAR_%28file_format%29
May be only one tip for the Total Commander - to enter the archive just use Ctrl-Pagedown. And I don't think that is something wrong on jar editing (in some circumstances you have no other option). However class files editing... it is a different game ...

JAR - Listing files into a folder

I would like to get a list of file contained in a directory which is in a jar package.
I have an "images" folder, within it I have an Images class that should load all images from that directory.
In the past i used the MyClass.class.getResourceAsStream("filename"); to read files, but how do I read a directory?
This is what I tried:
System.out.println(Images.class.getResource("").getPath());
System.out.println(new File(Images.class.getResource("").getPath()).listFiles());
I tried with Images.class.getResource because I have to work with File and there isn't a constructor that accepts an InputStream.
The code produces
file:/home/k55/Java/MyApp/dist/Package.jar!/MyApp/images/
null
So it is finding the folder which I want to list files from, but it is not able to list files.
I've read on other forums that in fact you can't use this method for folders in a jar archive, so how can I accomplish this?
Update: if possible, i would like to read files without having to use the ZipInputStream

You can't do that easily.
What you need to do:
Get the path of the jar file.
Images.class.getResource("/something/that/exists").getPath()
Strip "!/something/that/exists".
Use Zip File System to browse the Jar file.
It's a little bit of hacking.

read jar file variables from a configuration file

I have a java program which uses an external jar file. The jar file has some variables(IP addresses) that needs to be read from a configuration file.
How to I create one such configuration file? How to do I read it in jar?
[For now, I have hard-coded the needed variables data in the jar file. But I want to change it as and when I like, so changing in the configuration file is easier and I need not export(or re-compile) my jar file always]
To be more clear:
I have two java files: A.java, B.java.
"A.java" is to be exported as a jar file and used by "B.java".
Now, I have to read some data(settings, IP addresses) within A.java. All these sort of data have to kept in a separate file and should be read into A.java. After enabling this, A.java will be exported as a jar file and used by B.java.
How could this be achieved?

There are two different ways. If the jar is part of your classpath, then use Class.getResourceAsStream
It this is from plain jar some where, use JarFile

Getting XML file not in the same root as the source file in Java

I have a problem where I can't seem to link to a xml file, see the layout below:
Folder Name
-Folder
-Folder
-SourceFiles
-packagename
-all my java files
-myXml.xml
Build is where all the class files etc is stored.
src is where the projectFolder is, and within it the java files
Code I am using to link XML File for Synth: SynthDialog.class.getResourceAsStream("synthtest/synthDemo.xml")
Now I want to link to the myXML.xml file in the top-level folder. It would be the PHP Equivelent of ../../Folder/
Thanks

You appear to be attempting to access the file using getResourceAsStream with a relative name. If that is the case, then the resource should be in located in a JAR file or directory on the classpath, and the location will be resolved relative to the FQN of the class.
I can't tell where the ".class" files are located in the tree, or how your classpath is set up, so I can't be more specific.
UPDATED
If you are executing out of that build directory, then your build process needs to copy the XML file to the appropriate place in the build tree so that the class-relative path ends up referring to the file. (Or use a path that starts with "/" so that you don't depend on the classes FQN at all.)
In the long term, you will probably execute out of a JAR file, and the data file will need to be inside it.

Use "getSystemResourceAsStream" instead of "getResourceAsStream" to access files outside of your codebase.