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How to access outside file from java jar

I have a jar that is reading a file using below code:
Thread.currentThread().getContextClassLoader().getResource(fileName);
I want to run this jar using java -jar .jar command but I want to keep this file outside my jar, so that I can edit the jar file later on without touching the jar. Can anyone help me, how to run this jar so that it will pick up the file from outside.

There are multiple approaches you can do that and it will depend on where would you like to place this external file. For the sake of this answer, I will refer to this file as config file
Not In The Same Directory
The first approach is where you will need to place this file outside the JAR and not necessarily next to the JAR file in the same directory. In that case, you can pass the file location of the config file using either an environment variable (if you are running the JAR in a shell for example) or a Java property.
To use an environment variable, assuming you are using some Linux distro, then you can use the export command to set the value; something like this:
$ export CONFIG_FILE_LOC=/etc/myapp/config.file
You can then read the value in your code using the System class by using the following code:
String fileLocationEnv = System.getenv("CONFIG_FILE_LOC");
Alternatively, you can set this as a property by adding the following segment to your launch command:
$ java -Dconfig.file.location=/etc/myapp/config.file -jar myapp.jar
You can then read the value in your code using the System class for properties using the following code:
String fileLocationProp = System.getProperty("config.file.location");
In The Same Directory
If you need the config file to co-exist in the same directory as your JAR file, then you can use the following code to get the JAR directory and then append the filename to it. Here's the code (assuming a class named MyApp)
try{
new File(MyApp.class.getProtectionDomain().getCodeSource().getLocation().toURI());
} catch(URISyntaxException exception){
System.out.println("Exception");
}
Hope that helps.

To open the file as a resoure, add the folder containing the file(s) you want to use, to your classpath:
java -classpath .;config -jar myjar.jar
This example adds the current directory and the config directory to your classpath.
Multiple folders can be specified by using a separator. On windows use ';' , on unix use ':' .
To open the file as a File, you can just use
new File("configfile")
which will look in the working directory (directory where you launched your java)

File object not working in Java Web App

I have a simple java class in my web application in which i have written the below code but its not working
File test= new File("/templates/xmdForModel.xsd");
templates folder is inside the root folder of the application.
the location of the file is ----> application-root/package/test.java
location of the file is --------> application-root/testRoot/template/xmdForModel.xsd
Error
Failed to read schema document 'file:/templates/xmdForModel.xsd', because 1) could not find the document; 2) the document could not be read; 3) the root element of the document is not .

If you want to look up the file name for files inside of your web application, you can use ServletContext#getRealPath.
However, I would recommend loading your resources using the classloader with Class#getResourceAsStream. This way, it even works if the file does not really exist as a file (for example only inside of a jar).
If this is a file that a user is supposed to edit (or that you write to), I would place it outside of the web application, and then specify an absolute path (for example "/etc/myapp/conf/xmd.xsd") with a configurable prefix.

Path in the File constructor can be absolute or relative. When you start the path with '/' (in a linux based os), it is going to consider that path as an absolute path and create a file/folder in the root of your file structure (not in the root of the project). It will be like specifying c:\templates in windows machine.
Try removing the first slash and running your program. Removing first slash will make the part relative from your .java file. So you can use ../ to switch to parent folder.
java: application-root/package/test.java
file: application-root/testRoot/template/xmdForModel.xsd
So from your java file you will need to change directory to application root folder and then select template folder. Like following.
File x = new File("../testRoot/template/xmdForModel.xsd");
src:http://docs.oracle.com/javase/tutorial/essential/io/path.html
/home/sally/statusReport is an absolute path. All of the information
needed to locate the file is contained in the path string.
A relative path needs to be combined with another path in order to
access a file. For example, joe/foo is a relative path. Without more
information, a program cannot reliably locate the joe/foo directory in
the file system.

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.

How to specify relative file path in Java file so that it can still work after the file is put in jar file?

Suppose I have a Java class that needs to access a file with absolute path
/home/gem/projects/bar/resources/test.csv:
package com.example
class Foo {
String filePath = ????? // path to test.csv
String lines = FileInputStream(new File(filePath).readAllLines();
}
Where the path to Foo.java is /home/gem/projects/bar/src/com/example.
Of course I cannot specify absolute path to the resource file. This is because jar file will be distributed as library for any clients to use in their own environments.
Assume the resource file like test.csv is always in the same path relative to project root. When a jar is created containing Foo.class, this jar also contains test.csv in the same relative path ( relative to project root).
What is the way to specify relative path that would work no matter where the project bar is moved to? Also how can I create a jar file (which can be in any location) so that the path to the resource file test.csv would still be correct.
To keep things simple, I have used invalid Java API ( readAllLines() which reads all the lines and return a string containing entire file content. Also not using try/catch).
Assume csv file can be read as well as written to.
I hope this makes it clear now.

Put the test.csv file into the src folder and use this:
Foo.class.getResourceAsStream("/test.csv")
To get an InputStream for the file. This will work wherever the project is moved, including packaged as a JAR file.

Example:
ProjectX\src\Test.java
ProjectX\resources\config.properties
If you have the above structure and you want to use your config.properties file, this is how you do it:
InputStream input = new FileInputStream("./resources/config.projects");
In this example you don't have to worry about packaging your source into jar file. You can still modify your resources folder anytime.

Use getResource(), as shown here.

proper file path of a text file to read

I want to manipulate a file in my java program.The file to read must be paralled to my src folder.
What should I give as file path?

An elaborated example might help. From your question, what I get is,
Source Path : /home/user/project1/src/
File Path : /home/user/project1/src/
If this is the case, then once you build the project, the file path is not going to remain the same. So if you say that relative path for the file to open remains the same in built code, then you can use Class.getResourceAsStream(String path) which returns you the InputStream for given file. You can then construct the File object using it.
Refer this for details.

You should have a File object representing your src folder, and then create a new File object using that:
File textFile = new File(srcFolder, relativePath);
How you determine srcFolder really depends on the context.
EDIT: If you're just trying to read a file which is present at build time, you should include it in your built jar file and use either ClassLoader.getResourceAsStream or Class.getResourceAsStream to load it at execution time.
For example, if you have this structure:
src\
com\
xyz\
Foo.class
data\
input.txt
Then you could use Foo.class.getResourceAsStream("/data/input.txt") or Foo.class.getClassLoader.getResourceAsStream("data/input.txt"). Both will give you an InputStream you can use to load the data.