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Relative path using Level hierarchy(../../)

I want to use the relative path in xml files in our project. I have the files in the following location.
D:/SDC-Builds/SRDM2.3.0/SRDM/Svr/IdP/IdPserver/conf/attribute-r.xml
I have other xml file which needs to ref the above location, I use the following relative path to be independent of machines and folder names.
In D:/SDC-Builds/SRDM2.3.0/SRDM/Svr/IdP/IdPserver/others/service.xml, i am using the code like below
service.xml
<srv:ConfigurationResource="../../../../../../IdP/IdPserver/conf/attribute-r.xml">
</srv>
Please tell me am i using proper convention to refer the attribute-r.xml ?

If your Project Root Directory is SRDM, then you need to get back from your executable path.
Say you have your exe file at SRDM/Svr/bin/EXECUTABLE.exe then,
you need to mention in xml as
<srv:ConfigurationResource="../IdP/IdPserver/conf/attribute-r.xml"></srv>
ie. Current working directory is SRDM/Svr/bin/ and from that you need to get back up to common junction[Svr] in your case.

Correct Place to put .p12 File in a Java Web Application?

I am trying to access a .p12 in my java web application. Where is the correct place to put the file and what line of code do I need to access the file as an InputStream. I am using the line below, but the class is not finding the file.
I currently have the file in WEB-INF/classes/theFile and am using:
InputStream inputStream = this.getClass().getClassLoader().getResourceAsStream("nameOfTheFile.p12");
to access the file
Any help would be appreciated. Thanks!

This answer works for any file, not just ".p12" files.
Put a "/" in front of the name of your file (so, "/nameOfTheFile.p12" in your case) to specify an absolute path that will be resolved by using the classpath (which includes the WEB-INF/classes directory).
When you use a path without a leading "/", it is a relative path (vs. absolute path), specifically relative to the working directory of Tomcat or whatever container you are using. Even if you can find your resource using a relative path, you should not; doing so will end up hard coding deployment/packaging details within your application.

relative path in Java

My program has to use certain files that reside in another directory. Right now I am using absolute path to specify the location of the files. But I know that the relative position of these files will remain the same compared to where my program is residing. How I can use relative path in Java to specify the location of these files?
For example my program is residing in
/home/username/project/src/com/xyz/
..and the target files are in..
/home/username/project/secure/

For example my program is residing in /home/username/project/src/com/xyz/
and the target files are in /home/username/project/secure/
Knowing the place where your program's source code resides does not help. What you need to know is the current directory of the program when it is executed. That could be literally anything. Even if you are launching the application from (for example) Eclipse, the application launcher allows you to specify the "current directory" for the child process in the Run configuration.

Your current path.
currentPath= /home/username/project/src/com/xyz/;
Relative path to "/home/username/project/secure/" folder is
relativePath= ../../../secure;

In Java you can use getParentFile() to traverse up the tree.
File currentDir = new File(".");
File parentDir = currentDir.getParentFile();
This will be safe as you are not using system dependent file separator (../)

Java (maven web app), getting full file path for file in resources folder?

I'm working with a project that is setup using the standard Maven directory structure so I have a folder called "resources" and within this I have made a folder called "fonts" and then put a file in it. I need to pass in the full String file path (of a file that is located, within my project structure, at resources/fonts/somefont.ttf) to an object I am using, from a 3rd party library, as below, I have searched on this for a while but have become a bit confused as to the proper way to do this. I have tried as below but it isn't able to find it. I looked at using ResourceBundle but that seemed to involve making an actual File object when I just need the path to pass into a method like the one below (don't have the actual method call in front of me so just giving an example from my memory):
FontFactory.somemethod("resources/fonts/somefont.ttf");
I had thought there was a way, with a project with standard Maven directory structure to get a file from the resource folder without having to use the full relative path from the class / package. Any advice on this is greatly appreciated.
I don't want to use a hard-coded path since different developers who work on the project have different setups and I want to include this as part of the project so that they get it directly when they checkout the project source.
This is for a web application (Struts 1.3 app) and when I look into the exploded WAR file (which I am running the project off of through Tomcat), the file is at:
<Exploded war dir>/resources/fonts/somefont.ttf

Code:
import java.io.File;
import org.springframework.core.io.*;
public String getFontFilePath(String classpathRelativePath) {
Resource rsrc = new ClassPathResource(classpathRelativePath);
return rsrc.getFile().getAbsolutePath();
}
In your case, classpathRelativePath would be something like "/resources/fonts/somefont.ttf".

You can use the below mentioned to get the path of the file:
String fileName = "/filename.extension"; //use forward slash to recognize your file
String path = this.getClass().getResource(fileName).toString();
use/pass the path to your methods.

If your resources directory is in the root of your war, that means resources/fonts/somefont.ttf would be a "virtual path" where that file is available. You can get the "real path"--the absolute file system path--from the ServletContext. Note (in the docs) that this only works if the WAR is exploded. If your container runs the app from the war file without expanding it, this method won't work.

You can look up the answer to the question on similar lines which I had
Loading XML Files during Maven Test run
The answer given by BobG should work. Though you need to keep in mind that path for the resource file is relative to path of the current class. Both resources and java source files are in classpath

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.