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Path of the folder inside the jar file

I have jar file langdetect.jar.
It has a hierarchy shown in image
There is a class LanguageDetection at com/langdetect package.
I need to access the path of the profiles.sm folder from above class while executing the jar file.
Thanks in advance.

Jars are nothing else than Zip files and Java provides support for handling those.
Java 6 (and earlier)
You can open the jar file as a ZipFile and iterate over the entries of it. Each entry has a full path name inside the file, there is no such thing as relative path names. Though you have to take care, that all entries - although being absolute in the zip file - do not start with a '/', if you need this, you have to add it. The following snippet will get you the path of a class file. The className has to end with .class, i.e. LanguageDetection.class
String getPath(String jar, String className) throws IOException {
final ZipFile zf = new ZipFile(jar);
try {
for (ZipEntry ze : Collections.list(zf.entries())) {
final String path = ze.getName();
if (path.endsWith(className)) {
final StringBuilder buf = new StringBuilder(path);
buf.delete(path.lastIndexOf('/'), path.length()); //removes the name of the class to get the path only
if (!path.startsWith("/")) { //you may omit this part if leading / is not required
buf.insert(0, '/');
}
return buf.toString();
}
}
} finally {
zf.close();
}
return null;
}
Java 7/8
You may open the JAR file using the Java7 FileSystem support for JAR files. This allows you to operate on the jar file as if it would be normal FileSystem. So you could walk the fileTree until you have found your file and the get the Path from it. The following example uses Java8 Streams and Lambdas, a version for Java7 could be derived from this but would be a bit larger.
Path jarFile = ...;
Map<String, String> env = new HashMap<String, String>() {{
put("create", "false");
}};
try(FileSystem zipFs = newFileSystem(URI.create("jar:" + jarFileFile.toUri()), env)) {
Optional<Path> path = Files.walk(zipFs.getPath("/"))
.filter(p -> p.getFileName().toString().startsWith("LanguageDetection"))
.map(Path::getParent)
.findFirst();
path.ifPresent(System.out::println);
}
Your particular Problem
The above solutions are for finding the path inside a Jar or Zip, but may possibly not be the solution to your problem.
Im not sure, whether I understand your problem correctly. As far as I see it, you'd like to have access to the path inside the classfolder for any purpose. The problem with that is, that the Class/Resource lookup mechanism doesn't apply to folders, only files. The concept that is close is a package, but that is always bound to a class.
So you always need a concrete file to be accessed via getResource() method. For example MyClass.class.getResource(/path/to/resource.txt).
If the resources are located in a profiles.sm folder relative to a class and its package, i.e. in /com/languagedetect/profile.sm/ you could build the path from the reference class, for example the class LanguageDetection in that package and derive the absolute path from this to the profiles.sm path:
String basePath = "/" + LanguageDetection.class.getPackage().getName().replaceAll("\\.", "/") + "/profiles.sm/";
URL resource = LanguageDetection.class.getResource(basePath + "myResource.txt");
If there is only one profiles.sm in the root of the jar, simply go for
String basePath = "/profiles.sm/";
URL resource = LanguageDetection.class.getResource(basePath + "myResource.txt");
If you have multiple jars with a resource in /profiles.sm, you could gain access to all of those via the classloader and then extract the Jar file from the URL of the class
for(URL u : Collections.list(LanguageDetection.class.getClassLoader().getResources("/profiles.sm/yourResource"))){
System.out.println(u);
}
In any case it's not possible without accessing the zip/jar file to browse the contents of this path or folder because Java does not support browsing for classes or resources inside a package/folder in classpath. You may use the Reflections lib for that or extend the ClassLoader example above by additionally reading the content of the detected jars using the zip example from above.

How to read a file that i created inside my the same package?

This is a chunk of data I'd like to access by a method.
I'm doing the following to read my file:
String fileName = "file.txt"
InputStream inputStream = new FileInputStream(fileName);
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
My file.txt is in the same package, but I still get FileNotFoundException.
I didn't use a path url to point to the file because I thought since this it going to be an android application, hard-coding the path might not work when deployed... Please correct me if I am wrong. Thanks bunch!

This shows how to do that. https://stackoverflow.com/a/14377185/2801237
Also the 'package' your class is in has nothing to do with the 'path' where the file is being executed from. (two different concepts, 'package' = folder hierarchy of java source code files), 'path' = location on a filesystem of a specific file, your APK is being 'executed' in a particular place, and the location it writes a file is associated with that (I actually don't know where 'offhand' it writes by default, because I always get cache dir, or sd card root, etc.)

You may use:
InputStream inputStream = this.getClass().getResourceAsStream(fileName);

How to get absolute path of existing folder

I have a test data folder in my java project... something like this
Project
src Folder
testData Folder
How can I get the data folder using java, previously we have a hardcoded the in a constant, but now we are using a CI tool and I want to avoid the hardcoded path.

its is very fine to hardcode the testdata folder, but the
point is that it should be a relative path:
At the top of the test class you can just write
static final String TEST_PATH = "./testdata/";
Note that this is relative to the project.
If you want to convert such a relative path to an absoulte one:
String absoluteTestPath = new File(TEST_PATH).getAbsolutePath();

Take the input parameter and create a new File Object. The Java file object will provide getPath(), getAbsolutePath(), and getCanonicalPath(). Be careful with the file systems. If you are swapping between Windows/Linux/Mac/etc, you may have issues with translating paths between source and destination. I assume you source and destination are on the same file system in the example below.
File f = new File(pathStr);
String absPath = f.getAbsolutePath();

How about using Paths?
Paths.get("your-relative-path-here").toAbsolutePath()

Well, I solved my issue in the following way
public static final String Path_TestData = System.getProperty("user.dir") + "\\TestData\\";

For src
static String SRC_PATH = new File(".").getCanonicalPath()+System.getParameter("file.separator")+SRC;
// where SRC is the name of your source folder
For testdata:-
static String TEST_PATH = "./testdata/";

Getting error saying file won't open in Java...any idea why this is happening?

I think I am really close, but I am unable to open a file I have called LocalNews.txt. Error says can't find file specified.
String y = "LocalNews.txt";
FileInputStream fstream = new FileInputStream(y);
DataInputStream in = new DataInputStream(fstream);
BufferedReader br = new BufferedReader(new InputStreamReader(in));
Name of file is LocalNews.txt in library called News....anyone know why the file will not open?
The file is in the same Java Project that I am working on.
Error: LocalNews.txt (The system cannot find the file specified)
Project is named Bst, package is src in subPackage newsFinder, and library that the text files are stored in is called News.
Found out it was looking in
C:\EclipseIndigoWorkspace1\Bst\bin\LocalNews.txt
But I want it to look in (I believe)
C:\EclipseIndigoWorkspace1\Bst\News\LocalNews.txt
But if I make the above url a string, I get an error.

String y = "LocalNews.txt";
instead use
String y = "path from root/LocalNews.txt"; //I mean the complete path of the file

Your program can probably not find the file because it is looking in another folder.
Try using a absolute path like
String y = "c:\\temp\\LocalNews.txt";

By 'library called News' I assume you mean a jar file like News.jar which is on the classpath and contains the LocalNews.txt file you need. If this is the case, then you can get an InputStream for it by calling:
InputStream is = Thread.currentThread().getContextClassLoader()
.getResourceAsStream("LocalNews.txt");

Use
System.out.println(System.getProperty("user.dir") );
to find out what your current directory is. Then you'll know for sure whether your file is in the current directory or not. If it is not, then you have to specify the path so that it looks in the right directory.
Also, try this -
File file = new File (y);
System.out.println(file.getCanonicalPath());
This will tell you the exact path of your file on the system, provided your file is in the current directory. If it does not, then you know your file is not in the current directory.

Create File object of file from parent directory in java

I have this issue of accessing a file in one of the parent directories.
To explain, consider the following dir structure:-
C:/Workspace/Appl/src/org/abc/bm/TestFile.xml
C:/Workspace/Appl/src/org/abc/bm/tests/CheckTest.java
In the CheckTest.java I want to create a File instance for the TestFile.xml
public class Check {
public void checkMethod() {
File f = new File({filePath value I want to determine}, "TestFile.xml");
}
}
I tried a few things with getAbsolutePath() and the getParent() etc but was getting a bit complicated and frankly I think I messed it up.
The reason I don't want to use "C:/Workspace/Appl/src/org/abc/bm" while creating the File instance is because the C:/Workspace/Appl is not fixed and in all circumstances will be different at runtime and basically I don't want to hard-code.
What could be the easiest and cleaner way to achieve this ?
Thank you.

You should load it from Classpath in this case.
In your CheckTest.java, try
FileInputStream fileIs = new FileInputStream(CheckTest.class.getClassLoader().getResourceAsStream("org/abc/bm/TestFile.xml");

Use System.getProperty to get the base dir or you set the base.dir during application launch
java -Dbase.dir=c:\User\pkg
System.getProperty("base.dir");
and use
System.getProperty("file.separator");

What could be the easiest and cleaner way to achieve this ?
For accessing static resources use:
URL urlToResource = this.getClasS().getResource("path/to/the.resource");
If the resource is expected to change, write it to a sub-directory of user.home, where it is easy to locate later.

First of all, you can't get a reference to the source file path on runtime.
But, you can access the resrources included at your classpath (where you complied .class files will be).
Normally, your compiler will copy the xml file included at your srouce directory into the build directory, so at last, you could end up having something like this:
C:/Workspace/Appl/classes/org/abc/bm/TestFile.xml
C:/Workspace/Appl/classes/org/abc/bm/tests/CheckTest.class
Then, with your classpath pointing to the compiled classes root dir, you get the resources from this directory, using the ClassLoader.getResource method (or the equivalent Class.getResource() method).
public class Check {
public void checkMethod() {
java.net.URL fileURL=this.getClass().getResource("/org/abc/bm/tests/TestFile.xml");
File f=new File( fileURL.toURI());
}
}

One could do this:
String pathOfTheCurrentClass = this.getClass().getResource(".").getPath();
File file = new File(pathOfTheCurrentClass + "/..", "Testfile.xml");
or
String pathOfTheCurrentClass = this.getClass().getResource(".").getPath();
File filePath = new File(pathOfTheCurrentClass);
File file = new File(filePath.getParent(), "Testfile.xml");
But as Tomas Naros points out this gives you the file located in the build path.

Did you try
URL some=Test.class.getClass().getClassLoader().getResource("org/abc/bm/TestFile.xml");
File file = new File(some.getFile());