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File not found from resources folder

I have a file in my src/main/resources folder, the location of which i am passing in library function, but the file is not being read.
here is how the folder structure looks on decompiling the war
.
I have a file in my src/main/resources folder(Intuit.cto.gateway.aws.preprod.jks), the location of which i am passing in library function, but the file is not being read.
here is how the folder structure looks on decompiling the war.
when i create a simple maven project and try to do the same, i am able to access the file from src/main/resources
when i decompile my sample project it looks like this
To access the file, i am using this code :
public static void main(String args[])throws Exception{
FileReader fr=new FileReader("src/main/resources/filename");
int i;
while((i=fr.read())!=-1)
System.out.print((char)i);
fr.close();
}
how to solve this problem ?

You can't do it the way you want to.
Your path is most likely incorrect. Unzip your *.war file and see that src/main is no more.
If the war file is not exploded, you can't access its content by using simply path passed to a File* class, because your file is packaged as in zipped into standard location in a ZIP file with file extension WAR.
If your destination requires File or path which is passed to FileReader you have to read this file out of WAR file (using ClassLoader.getResourceAsStream()) and copying it out to a temporary location like File.createTempFile() or System.getProperty("java.io.tmpdir").

A resource (on the class path) is not necessarily a file system file. If the usage cannot deal with just an InputStream, one needs to create a temp file.
Path tempFile = Files.createTempFile("phoneid-", ".jks");
InputStream res = getClass().getResourceAsStream("/Intuit/gateway/preProd.jks");
// Case-sensitive, absolute path on the class path.
Files.copy(res, tempFile);
// Maybe:
//tempFile.toFile().deleteOnExit();
String param = tempFile.toString();
(A result null for getResource/getResourceAsStream indicates a wrong path.)
A partial solution would also be to check the URL of the resource, whether has the "file:" protocol, and not "jar:file:" (packed in a zip format, like jar, ear or war).
URL url getClass().getResource("/Intuit/gateway/preProd.jks");
With a class, the path can either be relative to the class package directory, or absolute ("/..."). If using the ClassLoader instead the path is always absolute, and should be written without / in front.

File only accessible when placed in a location where it will be erased from the jar

I'm new to NetBeans IDE, and am struggling with accessing a file after building the jar file. After reading through many posts on this topic, I decided to try the following code:
BufferedReader read = new BufferedReader(new InputStreamReader(getClass().getResourceAsStream("/file.txt")));
This works fine when my file is placed inside the "build" folder of the project where the .class files are, but of course this is a problem because it is erased in the "clean and build" process when the jar file is created. I have tried placing it in the src folder, in a separate "resources" package, and in the root of directory. I have also tried calling getResourceAsStream() with "file.txt" and "/src/file.txt," but it only works in the above configuration when the file is with the .class files. Any tips would be much appreciated!

Why not have your file folder inside the tomcat bin and refer the directory from your code. So maven clean will not alter the files and you can remove, update file without needing to restart the application. ( here i have file inside etc )
Path: /Users/username/Documents/apache-tomcat-8.5.15/bin/etc
ArrayList<String> readList = null;
String workingDir = System.getProperty("user.dir");
String fileName = "File.txt";
File file = new File(workingDir+"/etc/" + fileName);
readList = resourceReader.readFile(file.getAbsolutePath());
I have method readFile to parse some data and build the ArrayList in the above example.
Read about System Properties

Turns out the solution was really simple...I had been trying to manually create a resources folder, but the contents kept being deleted upon building of the jar. Instead, I created a resources package and put the file into the auto-generated folder inside the src folder, which packaged the file into the jar. Thanks everyone!

text file inside class folder, yet cannot find it

As you can see, the file is right inside the same folder where the class file is. While I tried to have it print the path to it, that is a bummer because no matter what name I type it will always type a full path to that invented file name. The file is comma separated, that I copied pasted into gedit and tried all kinds of extensions, csv, txt and no extension.
What I am trying to do is import it into an array, but the error is clear, no such file or directory.

You can try getting the path of the root folder from System.getProperty(). Then just append underlying folder names till the required filename.
private static String CLASS_PATH = System.getProperty("user.dir");
public static String FILE_PATH= CLASS_PATH + File.separator + "Source Packages"+ File.separator + "javaapplication2"+ File.separator + "pedidos.txt";

You should place files that are needed by your program in the classpath of your application so that they will get packaged in your JAR file. You can then read them with getResourceAsStream() Here's an idea of how that would work.
public class ArrayMain {
public static void main(String[] args) {
Scanner in = new Scanner(ArrayMain.class.getResourceAsStream("numbers.txt"));
String[] numbers = in.nextLine().split(",");
for (int i = 0; i < numbers.length; i++) {
System.out.println(numbers[i]);
}
}
}
The file numbers.txt is a simple file that looks like this:
1,2,3,4,5,6,7,8,9

Do not use relative paths to get files packaged with your application, it's dirty and senseless.
Path are relative to the folder in which was launched the java process. It's (usually) not your application source directory and never you file.java directory.
Even then, it's a terrible practice to put files (resources) inside your application source code. Because your java files needs to be compilated (and often packaged into a jar), the current architecture is likely to have no more meaning once packaged.
If you want to refer to a file on your computer, put it in a sensible folder (like /user/documents) and get it with its absolute path.
If you want the file to be packaged with your application, put it in a resource folder, package it in a jar and get the path with a classpath:// scheme.

The file needs to be placed at the root folder of your project.

Reading file right outside jar

So I have a .jar file in a folder and I have some input files in that folder. However, the program looks for the file in the home folder (several layers up). I want it obviously to read it from the folder that it's in but I don't want to be explicit about the file path to my folder because other people won't necessarily put their .jar file in the same spot.
Is there a way to read a file directly outside of the jar file? If not, is there a way to do this without hard-coding the file path?
edit:
here's the code. It just checks if the input files exist.
package main;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
/**
* Created by chris on 12/1/15.
* Control class that will be run when the jar is run.
*/
public class run {
public static void main(String[] args) throws Exception {
if (!(new File("settings.txt").exists())) {
start.run();
}
if (!(new File("api_key.txt").exists())) {
alert.display("Make your api_key.txt please.");
} else {
gatherData.run();
}
}
}
edit 2:
I've tried adding relative references with "./" at the beginning but that doesn't work either.

If you can rely on your .jar file being on the file system, you can get the absolute path of it by
new File(run.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
which can result in a SecurityException if a SecurityManager is present and not allowing this.
Another possibility would be to use
(new File(System.getProperty("java.class.path"))).getAbsolutePath();
The folder of the jar file can then be obtained using getParentFile()

If I am not wrong you are trying to access a file right in the same folder as that of the .jar file.
This can easily be done using the relative URL. By relative URL, I meant using
new File("./settings.txt"), this searches for the file in the folder same as that of the running .jar file. however you can use "../settings.txt" to look for the file one folder up.
"./" refers same directory "../" refers one directory up.

Just use ./ before file names.

The JAR files not neccesarly load from file system, and when loaded from file system can be any directory where the application start from - so relative path is not a good idea. The JAR files can from other types of source not only FileSystem, because it can be stream. I think the best way if there is a system parameter where you can pass the directory or working directory when you start the JAR file.

java.io.FileNotFoundException: the system cannot find the file specified

I have a file named "word.txt".
It is in the same directory as my java file.
But when I try to access it in the following code this file not found error occurs:
Exception in thread "main" java.io.FileNotFoundException: word.txt
(The system cannot find the file specified)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(Unknown Source)
at java.util.Scanner.<init>(Unknown Source)
at Hangman1.main(Hangman1.java:6)
Here's my code:
import java.io.File;
import java.util.*;
public class Hangman1 {
public static void main(String[] args) throws Exception {
Scanner input = new Scanner(new File("word.txt"));
String in = "";
in = input.nextLine();
}
}

Put the word.txt directly as a child of the project root folder and a peer of src
Project_Root
src
word.txt
Disclaimer: I'd like to explain why this works for this particular case and why it may not work for others.
Why it works:
When you use File or any of the other FileXxx variants, you are looking for a file on the file system relative to the "working directory". The working directory, can be described as this:
When you run from the command line
C:\EclipseWorkspace\ProjectRoot\bin > java com.mypackage.Hangman1
the working directory is C:\EclipseWorkspace\ProjectRoot\bin. With your IDE (at least all the ones I've worked with), the working directory is the ProjectRoot. So when the file is in the ProjectRoot, then using just the file name as the relative path is valid, because it is at the root of the working directory.
Similarly, if this was your project structure ProjectRoot\src\word.txt, then the path "src/word.txt" would be valid.
Why it May not Work
For one, the working directory could always change. For instance, running the code from the command line like in the example above, the working directory is the bin. So in this case it will fail, as there is not bin\word.txt
Secondly, if you were to export this project into a jar, and the file was configured to be included in the jar, it would also fail, as the path will no longer be valid either.
That being said, you need to determine if the file is to be an embedded-resource (or just "resource" - terms which sometimes I'll use interchangeably). If so, then you will want to build the file into the classpath, and access it via an URL. First thing you would need to do (in this particular) case is make sure that the file get built into the classpath. With the file in the project root, you must configure the build to include the file. But if you put the file in the src or in some directory below, then the default build should put it into the class path.
You can access classpath resource in a number of ways. You can make use of the Class class, which has getResourceXxx method, from which you use to obtain classpath resources.
For example, if you changed your project structure to ProjectRoot\src\resources\word.txt, you could use this:
InputStream is = Hangman1.class.getResourceAsStream("/resources/word.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
getResourceAsStream returns an InputStream, but obtains an URL under the hood. Alternatively, you could get an URL if that's what you need. getResource() will return an URL
For Maven users, where the directory structure is like src/main/resources, the contents of the resources folder is put at the root of the classpath. So if you have a file in there, then you would only use getResourceAsStream("/thefile.txt")

Relative paths can be used, but they can be tricky. The best solution is to know where your files are being saved, that is, print the folder:
import java.io.File;
import java.util.*;
public class Hangman1 {
public static void main(String[] args) throws Exception {
File myFile = new File("word.txt");
System.out.println("Attempting to read from file in: "+myFile.getCanonicalPath());
Scanner input = new Scanner(myFile);
String in = "";
in = input.nextLine();
}
}
This code should print the folder where it is looking for. Place the file there and you'll be good to go.

Your file should directly be under the project folder, and not inside any other sub-folder.
If the folder of your project is named for e.g. AProject, it should be in the same place as your src folder.
Aproject
src
word.txt

Try to create a file using the code, so you will get to know the path of the file where the system create
File test=new File("check.txt");
if (test.createNewFile()) {
System.out.println("File created: " + test.getName());
}

I was reading path from a properties file and didn't mention there was a space in the end.
Make sure you don't have one.

Make sure when you create a txt file you don't type in the name "name.txt", just type in "name". If you type "name.txt" Eclipse will see it as "name.txt.txt". This solved it for me. Also save the file in the src folder, not the folder were the .java resides, one folder up.

I have the same problem, but you know why? because I didn't put .txt in the end of my File and so it was File not a textFile, you shoud do just two things:
Put your Text File in the Root Directory (e.x if you have a project called HelloWorld, just right-click on the HelloWorld file in the package Directory and create File
Save as that File with any name that you want but with a .txt in the end of that
I guess your problem is solved, but I write it to other peoples know that.
Thanks.

i think it always boils to the classpath. having said that if you run from the same folder where your .class is then change Scanner input = new Scanner(new File("word.txt")); to Scanner input = new Scanner(new File("./word.txt")); that should work