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Nested jar: URI

The format of the jar: URI scheme is simply defined to be jar:<url>!/[<entry>].
Is it possible to "nest" two such URIs together and refer to a file inside an archive inside an archive this way? Something like:
jar:jar:http://example.com/!/archive!/file
This should designate /file inside an archive jar:http://example.com/!/archive, that is /archive found in a file provided by http://example.com/. However, URL.openConnection throws an exception:
java.net.MalformedURLException: no !/ in spec
I also cannot replace ! with %21 since that makes the jar: URI invalid, and I am not aware of any escaping that could be performed. Is this somehow possible? Or, as a more generic question, is it possible to store a URI with !/ inside <url> so that it is preserved?

As it is stated in the source code java.net.JarURLConnection does not handle nested Jar URIs.
Your example jar:jar:http://example.com/!/archive!/fileis parsed as entry archive!/file in the archive jar:http://example.com/. The latter is not a valid jar URI. You can not either escape the !/ sequences, because the text after jar: and before !/ must be a valid URI without any translation.
You can of course provide your own URLConnection implementation that supports nested Jar URIs.

While the other answer is indeed correct in the case of JarURLConnection, the use of jar: and similar URIs is not restricted to it. Such is the case of the Apache Commons VFS:
jar:// arch-file-uri[! absolute-path]
Where arch-file-uri refers to a file of any supported type, including other zip files. Note: if you would like to use the ! as normal character it must be escaped using %21.
"Normal character" means usage of the character in path to the file inside the archive, not in the inner URI. This makes jar:jar:http://example.com/!/archive!/file valid, as it is the last occurence of ! that should be used as a delimiter.
Note that the inner URI should probably still remain unescaped, thus it is impossible to use # there.

File wildcard use *

I am trying to read a file which has name: K2ssal.timestamp.
I want to handle the time stamp part of the file name as wildcard.
How can I achieve this ?
tried * after file name but not working.
var getK2SSal: Iterator[String] = Source.fromFile("C://Users/nrakhad/Desktop/Work/Data stage migration/Input files/K2Ssal.*").getLines()

You can use Files.newDirectoryStream with directory + glob:
import java.nio.file.{Paths, Files}
val yourFile = Files.newDirectoryStream(
Paths.get("/path/to/the/directory"), // where is the file?
"K2Ssal.*" // glob of the file name
).iterator.next // get first match

Misconception on your end: unless the library call is specifically implemented to do so, using a wildcard simply doesn't work like you expect it to.
Meaning: a file system doesn't know about wildcards. It only knows about existing files and folders. The fact that you can put * on certain commands, and that the wildcard is replaced with file names is a property of the tool(s) you are using. And most often, programming APIs that allow you to query the file system do not include that special wild card handling.
In other words: there is no sense in adding that asterisk like that.
You have to step back and write code that actively searches for files itself. Here are some examples for scala.

You can read the directory and filter on files based upon the string.
val l = new File("""C://Users/nrakhad/Desktop/Work/Data stage migration/Input files/""").listFiles
val s = l.filter(_.toString.contains("K2Ssal."))

Intershop building url and access language property file from a pipeline

I am trying to create a JSON response but I need some things that is easily done in ISML. First I would like to know how I can build a proper URL this should link to a pipeline call with some parameters. The other thing I would like to know is how can I access language property files in a pipelet/java like the istext tag that you can call with ISML
Thank you

If you do not like to use the URL with the pipeline name, you can create a short URL in the back office, or create a URL rewrite rule.
Every parameter you add to the URL will of course be available in the pipeline dictionary.
To get a localized text in Java take a look at the GetLocalizedTextByKey pipelet.
It uses the LocalizationProvider to provide a localized text for a specific key and a locale.
You can also take a look at this Intershop Knowledge Base article if you have access: https://support.intershop.com/kb/index.php/Display/2258M5

Merging two relative file URLs

Let' say I have two paths, first can look like folder/ and second like /anotherFolder/image.png. I would like to merge those two paths in some automated fashion and with option for user to omit the last slash in first string and first slash in second string. So all of these
folder/ + /anotherFolder/image.png
folder + anotherFolder/image.png
folder + /anotherFolder/image.png
should give me folder/anotherFolder/image.png
I need to merge two properties in one of my projects and I want it as dummy as possible:)So is there some trick with URL class or do I have to play around with Strings?

You can do this with java.io.File, by using the constructor which takes a File and a String as arguments, will interpret the String as a relative path to the File.
Or with java.net.URL, you can send an URL and a String to the constructur, which will interpret the URL as a context for the String parameter.

I actually used FileUtils.getFile() from Apache Commons IO but Rolf's solution was working too.

Getting the inputstream from a classpath resource (XML file)

In Java web application, Suppose if I want to get the InputStream of an XML file, which is placed in the CLASSPATH (i.e. inside the sources folder), how do I do it?

ClassLoader.getResourceAsStream().
As stated in the comment below, if you are in a multi-ClassLoader environment (such as unit testing, webapps, etc.) you may need to use Thread.currentThread().getContextClassLoader(). See http://stackoverflow.com/questions/2308188/getresourceasstream-vs-fileinputstream/2308388#comment21307593_2308388.

ClassLoader.class.getResourceAsStream("/path/file.ext");

That depends on where exactly the XML file is. Is it in the sources folder (in the "default package" or the "root") or in the same folder as the class?
In for former case, you must use "/file.xml" (note the leading slash) to find the file and it doesn't matter which class you use to try to locate it.
If the XML file is next to some class, SomeClass.class.getResourceAsStream() with just the filename is the way to go.

ClassLoader.class.getResourceAsStream("/path/to/your/xml") and make sure that your compile script is copying the xml file to where in your CLASSPATH.

someClassWithinYourSourceDir.getClass().getResourceAsStream();

Some of the "getResourceAsStream()" options in this answer didn't work for me, but this one did:
SomeClassWithinYourSourceDir.class.getClassLoader().getResourceAsStream("yourResource");

I tried proposed solution and forward slash in the file name did not work for me, example: ...().getResourceAsStream("/my.properties"); null was returned
Removing the slash worked: ....getResourceAsStream("my.properties");
Here is from doc API:
Before delegation, an absolute resource name is constructed from the given resource name using this algorithm:
If the name begins with a '/' ('\u002f'), then the absolute name of the resource is the portion of the name following the '/'.
Otherwise, the absolute name is of the following form:
modified_package_name/name
Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e').