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Java listFiles in directory in jar

Is there any way to use listFiles() on a directory that's been packaged into a jar?
Let's say I have a directory in my resource directory with some text files: texts/text1.txt and texts/text2.txt.
And within this Java program I have a class that needs to use listFiles() to get a list of those files. I'll get something like jar:file:/home/soupkitchen.jar/!text. I'd expect that to be a directory. Is there any way to be able to treat it as a java.io.File directory containing files? Right now it seems to only be listed as neither a file nor directory.

No. java.io.File can only be used to list real directories.
However, you can treat it as a java.nio.file.Path.
Overall, you have three options:
Open the .jar as a Zip File System and use Files.newDirectoryStream or Files.list.
Iterate through all entries in the .jar file, looking for names that match.
Put a text file in your .jar that contains the names of all the entries in the directory, so you don't have to try to list them.

How to access the content of the jar during runtime?

How can I access the content of the jar file which has been started. I want to create a big jar file which contains everything I need and then during runtime I want to copy some files of my jar into an external folder. Is this possible?

You want this.getClass().getClassLoader().getResourceAsStream. Example:
InputStream config =
this.getClass().getClassLoader().getResourceAsStream("config.txt");

The files in the JAR are not accessible as files, so you must use getResourceAsStream to read them. See access files and folders in executable jars how to access the files within the jar.
Following, you use the inputstream to write the files onto the file system.
See:
Easy way to write contents of a Java InputStream to an OutputStream
http://www.mkyong.com/java/how-to-convert-inputstream-to-file-in-java/

You can access any file via the ClassPath using the classloader. Start with Class.getResourceAsStream

Accessing documents packaged in a jar file

Im trying to get the file path of a document that is packaged as a resource in a jar file so that i can display it in a swing application. The way I have it now works when I run it from eclipse but if I export it to a runnable jar file I can't access the the documents that are packaged in the jar file. How can I get the file path of the document when its inside the jar file?
Here is the line of code showing how I am trying to access the document:
File document = new File(getClass().getResource("/resources/documents/document.pdf").getPath());

The only kind of "file path" that exists for something inside a JAR file is the path relative to the root of the JAR. But in your case it seems that you know it already (it's "/resources/documents/document.pdf"). Files inside a JAR file have no path that you can use to access them directly as they don't exist within the real file system. You need to use either getResource() or getResourceAsStream() to access them. I don't remember right now which classes are used for images in Swing, but look closely at those classes - they should have overloaded methods that accept something like InputStream or URL instead of file path.

Open file; try filesystem first, then JARs

I'm trying to have my application load a resource (binary file) transparently:
If the file exists under the current directory, open it.
If not, try looking in the current JAR file if applicable.
If not, try looking in other JAR files. (This is optional and I don't mind explicitly specifying which JAR files.)
So far I know of File which opens a local file and ClassLoader which has getResource* for JAR contents.
Is there a class which combines the two? If not, how should I go about writing it myself? Should I write a ClassLoader which also checks the local filesystem? Using File? (I'm very unfamiliar with Java and don't even know what's a good type to return. InputStream?)
Thanks
P.S. By "file" I mean "path", e.g. "data/texture1.png".

Doing #1 and #3 is pretty easy. Doing #2 (just looking in the current JAR only) is much harder as it requires you figuring out what JAR you
If you wanted to check the filesystem first, otherwise load from classpath, it would be something like:
public java.io.InputStream loadByName(String name) {
java.io.File f = new java.io.File(name);
if (f.isFile()) {
return new FileInputStream(f);
} else {
return getClass().getResource(name);
}
}
If you want to prefer loading from the same JAR file first, you will need to figure out where it is. Check out Determine which JAR file a class is from for more info on figuring out the JAR file you want to load the resource from.

A URLClassLoader should be able to load both and try the file path first if the file path is on the class path ahead of the jar.
Regarding your comments:
I know that relative jar URLs don't
work. That's why the Spring guys came
up with the Resource abstraction.
Read about it here.
You might want to check the answers
to this Question: Loading a file
relative to the executing jar
file. The problem is similar to
yours.

Current jar file and current directory are not concepts in the JVM like they are when you're running a shell script. You would need to specify a directory to be used for loading the files that you're interested in, such as with a system property while executing the JVM:
java -Ddirectory.to.scan=/home/aib
Then retrieve this property:
String dir = System.getProperty("directory.to.scan");
Now when talking about JAR files, all JAR files specified explicitly on the classpath when you start the JVM are loaded by the ClassLoader. You can get the ClassLoader of a specific class by:
InputStream is = <Your class>.class.getClassLoader().getResourceAsStream("binary file");
Note that any jar file loaded by the current class loader is searched.

Absolute Path of Project's folder in Java

Lots of confusion in this topic. Several Questions have been asked. Things still seem unclear.
ClassLoader, Absolute File Paths etc etc
Suppose I have a project directory structure as,
MyProject--
--dist
--lib
--src
--test
I have a resource say "txtfile.txt" in "lib/txt" directory. I want to access it in a system independent way. I need the absolute path of the project.
So I can code the path as abspath+"/lib/Dictionary/txtfile.txt"
Suppose I do this
java.io.File file = new java.io.File(""); //Dummy file
String abspath=file.getAbsolutePath();
I get the current working directory which is not necessarily project root.
Suppose I execute the final 'prj.jar' from the 'dist' folder which also contains "lib/txt/txtfile.txt" directory structure and resource,It should work here too. I should absolute path of dist folder.
Hope the problem is clear.

You should really be using getResource() or getResourceAsStream() using your class loader for this sort of thing. In particular, these methods use your ClassLoader to determine the search context for resources within your project.
Specify something like getClass().getResource("lib/txtfile.txt") in order to pick up the text file.
To clarify: instead of thinking about how to get the path of the resource you ought to be thinking about getting the resource -- in this case a file in a directory somewhere (possibly inside your JAR). It's not necessary to know some absolute path in this case, only some URL to get at the file, and the ClassLoader will return this URL for you. If you want to open a stream to the file you can do this directly without messing around with a URL using getResourceAsStream.
The resources you're trying to access through the ClassLoader need to be on the Class-Path (configured in the Manifest of your JAR file). This is critical! The ClassLoader uses the Class-Path to find the resources, so if you don't provide enough context in the Class-Path it won't be able to find anything. If you add . the ClassLoader should resolve anything inside or outside of the JAR depending on how you refer to the resource, though you can certainly be more specific.
Referring to the resource prefixed with a . will cause the ClassLoader to also look for files outside of the JAR, while not prefixing the resource path with a period will direct the ClassLoader to look only inside the JAR file.
That means if you have some file inside the JAR in a directory lib with name foo.txt and you want to get the resource then you'd run getResource("lib/foo.txt");
If the same resource were outside the JAR you'd run getResource("./lib/foo.txt");

First, make sure the lib directory is in your classpath. You can do this by adding the command line parameter in your startup script:
$JAVA_HOME/bin/java -classpath .:lib com.example.MyMainClass
save this as MyProject/start.sh or any os dependent script.
Then you can access the textfile.txt (as rightly mentioned by Mark) as:
// if you want this as a File
URL res = getClass().getClassLoader().getResource("text/textfile.txt");
File f = new File(res.getFile());
// As InputStream
InputStream in = getClass().getClassLoader()
.getResourceAsStream("text/textfile.txt");

#Mark is correct. That is by far the simplest and most robust approach.
However, if you really have to have a File, then your best bet is to try the following:
turn the contents of the System property "java.class.path" into a list of pathnames,
identify the JAR pathname in the list based on its filename,
figure out what "../.." is relative to the JAR pathname to give you the "project" directory, and
build your target path relative to the project directory.
Another alternative is to embed the project directory name in a wrapper script and set it as a system property using a -D option. It is also possible to have a wrapper script figure out its own absolute pathname; e.g. using whence.