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How to have my java project to use some files without using their absolute path?

I have written a project where some images are used for the application's appearance and some text files will get created and deleted along the process. I only used the absolute path of all used files in order to see how the project would work, and now that it is finished I want to send it to someone else. so what I'm asking for is that how I can link those files to the project so that the other person doesn't have to set those absolute paths relative to their computer. something like, turning the final jar file with necessary files into a zip file and then that the person extracts the zip file and imports jar file, when runs it, the program work without any problems.
by the way, I add the images using ImageIcon class.
I'm using eclipse.

For files that you just want to read, such as images used in your app's icons:
Ship them the same way you ship your class files: In your jar or jmod file.
Use YourClassName.class.getResource or .getResourceAsStream to read these. They are not files, any APIs that need a File object can't work. Don't use those APIs (they are bad) - good APIs take a URI, URL, or InputStream, which works fine with this.
Example:
package com.foo;
public class MyMainApp {
public void example() {
Image image = new Image(MyMainApp.class.getResource("img/send.png");
}
public void example2() throws IOException {
try (var raw = MyMainApp.class.getResourceAsStream("/data/countries.txt")) {
BufferedReader in = new BufferedReader(
new InputStreamReader(raw, StandardCharsets.UTF_8));
for (String line = in.readLine(); line != null; line = in.readLine()) {
// do something with each country
}
}
}
}
This class file will end up in your jar as /com/foo/MyMainApp.class. That same jar file should also contain /com/foo/img/send.png and /data/countries.txt. (Note how starting the string argument you pass to getResource(AsStream) can start with a slash or not, which controls whether it's relative to the location of the class or to the root of the jar. Your choice as to what you find nicer).
For files that your app will create / update:
This shouldn't be anywhere near where your jar file is. That's late 80s/silly windows thinking. Applications are (or should be!) in places that you that that app cannot write to. In general the installation directory of an application is a read-only affair, and most certainly should not be containing a user's documents. These should be in the 'user home' or possibly in e.g. `My Documents'.
Example:
public void save() throws IOException {
Path p = Paths.get(System.getProperty("user.home"), "navids-app.save");
// save to that file.
}

Processing/Java File Count Issue With File Pathway (Variable Type)

Although the Title isn't very understandable I do have a simple issue. So i'm trying to write some code in a Processing Sketch (https://processing.org/) which can count how many files are in a document. The problem is, is that it doesn't accept the variable type.
File folder = File("My File Path");
folder.listFiles().size;
It says the function File(String) doesn't exist. When I try to put the file path without quation marks, it still doesn't work!
If you have a solution then please use a functioning example so that I know how it works. Thanks for any help!

As Joakim Danielson says it is constructor so you need to use new keyword.
Below code will work for you.
File folder = new File("My File Path");
int fileLength = folder.listFiles().length;

It's a constructor so you need to use new
File folder = new File("My File Path");
//To get the number of files in the folder
folder.listFiles().length;

Assuming the "My File Path" folder is inside your sketch you need to provide the path to your sketch. Luckily Processing already provides a helper function: sketchPath()
Here's an example:
File folder = new File(sketchPath("My File Path"));
println("folder.exists: " + folder.exists());
if(folder.exists()){
println(folder.listFiles().length + " files and/or directories");
}else{
println("folder does not exist, double check the path");
}
Bare in mind there's also a dataPath() function which points to a folder named data in your sketch folder. The data folder is typically used for storing external data (e.g. assets (raster or vector images/Processing font files) or raw data (binary/text/csv/xml/json/etc.)). This is useful to separate your sketch source files from the data to be loaded/accessed by your sketch.
Also, Processing has a few utility functions for listing files and folders.
Be sure to check out Processing > Examples > Topics > File IO > DirectoryList
The example includes less documented functions such as listFiles() (which returns an array of java.io.File objects based on the filters set) or listPaths (which returns an array of String objects: just the paths).
The options and filters are quite handy, for example if you want to list directories only and ignore files you can simply write simply like:
println("directories: " + listFiles(sketchPath("My File Path"),"directories").length);
For example if want to list all the wav files in a data/audio directory inside the sketch you can use:
File[] files = listFiles(dataPath("audio"), "files", "extension=wav");
This will ignore directories and any other file that does not have .wav extension.
To make this answer complete, here are a few more details on the options for listFiles/listPaths from Processing's source code:
"relative" -> no effect with the Files version, but important for listPaths
"recursive"-> traverse nested directories
"extension=js" or "extensions=js|csv|txt" (no dot)
"directories" -> only directories
"files" -> only files
"hidden" -> include hidden files (prefixed with .) disabled by default

How to get to a file by passing relative path in java?

I am trying to understand "How to get to file by passing relative path of a file or folder?" . Here is the example:
CODE:
public class somex {
public static void main {
String fileName = System.getProperty("user.dir"); <---This gives me path for the current working directory.
File file = new File(fileName + "../../xml_tutorial/sample.xlsx" );
System.out.println(file.getCanonicalPath()); <---This gives me path for the file that is residing in folder called "xml_tutorial".
}
}
>>>>
Here, I know the file location so i was able to pass correct relative path. And, managed to print the file path. I have deleted the "sample.xlsx" and executed the above code; With no failing it gives me the path name and it is same path as when the file exists (i.e. before deleting). How it is possible ? I am expecting EXCEPTION here. why it is not throwing exception ?
Two, I want to use regular expression for the file name, such as: "../../xml_tutorial/samp.*". But this doesn't do the job and it gives me IOException. Why it is not able to identify the file sample.xlsx ? (NOTE: this is when the file exist and one hundred precent sure there is only one file with the name "sample.xlsx")

I have deleted the "sample.xlsx" and executed the above code; With no failing it gives me the path name and it is same path as when the file exists (i.e. before deleting). How it is possible ? I am expecting EXCEPTION here. why it is not throwing exception ?
File doesn't care whether the file actually exists. It just resolves the path. There's no need for the file to exist in order to take the path
/home/tjc/a/b/c/../../file.txt
...and turn it into the canonical form
/home/tjc/a/file.txt
If you want to know whether the file on that path actually exists, you can use the exists() method.
On your second, unrelated question:
Two, I want to use regular expression for the file name, such as: "../../xml_tutorial/samp.*". But this doesn't do the job and it gives me IOException. Why it is not able to identify the file sample.xlsx ?
There's nothing in the File documentation saying that it supports wildcards. If you want to do searches, you'll want to use list(FilenameFilter) or listFiles(FilenameFilter) and a FilenameFilter implementation, or listFiles(FileFilter) and a FileFilter implementation.

Reading .txt file from another directory

The code I am running is in /Test1/Example. If I need to read a .txt file in /Test1 how do I get Java to go back 1 level in the directory tree, and then read my .txt file
I have searched/googled and have not been able to find a way to read files in a different location.
I am running a java script in an .htm file located at /Test1/Test2/testing.htm. Where it says script src=" ". What would I put in the quotations to have it read from my file located at /Test1/example.txt.

In Java you can use getParentFile() to traverse up the tree. So you started your program in /Test1/Example directory. And you want to write your new file as /Test1/Example.txt
File currentDir = new File(".");
File parentDir = currentDir.getParentFile();
File newFile = new File(parentDir,"Example.txt");;
Obviously there are multiple ways to do this.

You should be able to use the parent directory reference of "../"
You may need to do checks on the OS to determine which directory separation you should be using ['\' compared to '/']

When you create a File object in Java, you can give it a pathname. You can either use an absolute pathname or a relative one. Using absolutes to do what you want would require:
File file = new File("/Test1/myFile.txt");
if(file.canRead())
{
// read file here
}
Using relatives paths if you want to run from the location /Test1/Example:
File file = new File("../myFile.txt");
if(file.canRead())
{
// read file here
}

I had a similar experience.
My requirement is: I have a file named "sample.json" under a directory "input", I have my java file named "JsonRead.java" under a directory "testcase". So, the entire folder structure will be like untitled/splunkAutomation/src and under this I have folders input, testcase.
once after you compile your program, you can see a input file copy named "sample.json" under a folder named "out/production/yourparentfolderabovesrc/input" and class file named "JsonRead.class" under a folder named "out/production/yourparentfolderabovesrc/testcase". So, during run time, Java will actually refer these files and NOT our actual .java file under "src".
So, my JsonRead.java looked like this,
package testcase;
import java.io.*;
import org.json.simple.JSONObject;
public class JsonRead{
public static void main(String[] args){
java.net.URL fileURL=JsonRead.class.getClass().getResource("/input/sample.json");
System.out.println("fileURL: "+fileURL);
File f = new File(fileURL.toURI());
System.out.println("fileIs: "+f);
}
}
This will give you the output like,
fileURL: file:/C:/Users/asanthal/untitled/out/production/splunkAutomation/input/sample.json
fileIs: C:\Users\asanthal\untitled\out\production\splunkAutomation\input\sample.json

It worked for me. I was saving all my classes on a folder but I needed to read an input file from the parent directory of my classes folder. This did the job.
String FileName = "Example.txt";
File parentDir = new File(".."); // New file (parent file ..)
File newFile = new File(parentDir,fileName); //open the file

How to pass a text file as a argument?

Im trying to write a program to read a text file through args but when i run it, it always says the file can't be found even though i placed it inside the same folder as the main.java that im running.
Does anyone know the solution to my problem or a better way of reading a text file?

Do not use relative paths in java.io.File.
It will become relative to the current working directory which is dependent on the way how you run the application which in turn is not controllable from inside your application. It will only lead to portability trouble. If you run it from inside Eclipse, the path will be relative to /path/to/eclipse/workspace/projectname. If you run it from inside command console, it will be relative to currently opened folder (even though when you run the code by absolute path!). If you run it by doubleclicking the JAR, it will be relative to the root folder of the JAR. If you run it in a webserver, it will be relative to the /path/to/webserver/binaries. Etcetera.
Always use absolute paths in java.io.File, no excuses.
For best portability and less headache with absolute paths, just place the file in a path covered by the runtime classpath (or add its path to the runtime classpath). This way you can get the file by Class#getResource() or its content by Class#getResourceAsStream(). If it's in the same folder (package) as your current class, then it's already in the classpath. To access it, just do:
public MyClass() {
URL url = getClass().getResource("filename.txt");
File file = new File(url.getPath());
InputStream input = new FileInputStream(file);
// ...
}
or
public MyClass() {
InputStream input = getClass().getResourceAsStream("filename.txt");
// ...
}

Try giving an absolute path to the filename.
Also, post the code so that we can see what exactly you're trying.

When you are opening a file with a relative file name in Java (and in general) it opens it relative to the working directory.
you can find the current working directory of your process using
String workindDir = new File(".").getAbsoultePath()
Make sure you are running your program from the correct directory (or change the file name so that it will be relative to where you are running it from).

If you're using Eclipse (or a similar IDE), the problem arises from the fact that your program is run from a few directories above where the actual source is located. Try moving your file up a level or two in the project tree.
Check out this question for more detail.

The simplest solution is to create a new file, then see where the output file is. That is the correct place to put your input file into.

If you put the file and the class working with it under same package can you use this:
Class A {
void readFile (String fileName) {
Url tmp = A.class.getResource (fileName);
// Or Url tmp = this.getClass().getResource (fileName);
File tmpFile = File (tmp);
if (tmpFile.exists())
System.out.print("I found the file.")
}
}
It will help if you read about classloaders.

say I have a text file input.txt which is located on the desktop
and input.txt has the following content
i came
i saw
i left
and below is the java code for reading that text file
public class ReadInputFromTextFile {
public static void main(String[] args) throws Exception
{
File file = new File(
"/Users/viveksingh/desktop/input.txt");
BufferedReader br
= new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
output on the console:
i came
i saw
i left