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How do you create a file within a Java artifact?

I am currently working on a Java application in Intellij, and I cannot create a file within my artifact. As an example, I'm using File to create a file within the source, which is MainMenuData.txt.
File mainMenu = new File("MainMenuData.txt");
String absPath = mainMenu.getPath();
mainMenu.createNewFile();
BufferedReader br = new BufferedReader(new FileReader(absPath));
In this, I'm using File to make sure that file exists whenever it isn't.
Instead, I'd like to build within the (Production) artifact. Is that doable?
Anything helps. Thanks.

You could do that.
I assume your artifact is a jar file, which is nothing else than a zip file. You can obtain the location of your jar file in the file system and use the Zip File System to modify it. However I'm not sure, if the jvm might have a problem with that and it might not work on windows, since windows likes to block files, that are in use.
Also this would definitely fail, if your jar file is not stored locally.
A better approach would be to store your data files at the appropriate location in your system.
On Linux(and probably mac): <home>/.local/share/<your application name>/
On Linux(global): /var/lib/<application name>
On Windows(I think, better check it yourself): <appdata>/<your application name>
Your code would look something like this(for Linux):
File home = new File(System.getProperty("HOME");
File configDirectory = new File(home, ".local/share/<application name>");
configDirectory.mkdirs();
File mainMenu = new File(configDirectory, "MainMenuData.txt");
For windows do something similar. If you need both, you should check on which you are currently running.

NPE while try to get Resource in jar

I have tried many variants but I cant find correct.
I have something like
Inside my jar, which created by Maven I can see that
That is my folder with classes. And, by the way, If I start my program in IDEA, not from Console, there is not any exception with paths
Here, I am in debug mode start my jar trying to see, where is the problem.
If I do 'file.exists()' it would be false but file inside. I think, that problem because of '.jar!\' in the path, but I don`t know how to remove that.
Anyway I've tried absolute and relative path, I've tried
Thread.getCurrentThread.getContextLoader.getResource()
GUI.class.getResource()
GUI.class.getClassLoader.getResource()
Nothing help

You can't use File to open resources inside a jar file. File can only be used with normal files and directories.
Note: using File works fine within in the IDE, since all files are not packaged in a jar file yet. But the program will break after you package it.
Once you locate the resource eg. URL res = GUI.class.getResource("/rxtx64/myres.dll") , you can open that resource as a stream InputStream is = res.openStream(); .
See also related answers Utils to read resource text file to String (Java) and How to read a text-file resource into Java unit test?

FileInputStream in Spring MVC fails to find file

I am working on a SpringMVC project which runs a number of automated tests on a database. The access details for this database are located in a .properties file. This file is located within the project directory.
FileInputStream fis = new FileInputStream("batch-dm.properties");
propFile = new Properties();
propFile.load(fis);
As the file is stored in the project directory the FileInputStream should be able to access it no?
If I provide the absolute path e.g.
FileInputStream fis = new FileInputStream("C:/workspace/Project/batch-dm.properties");
It recognises the file and runs properly.
Do I need to sore this file in a different location because it is a Spring MVC project?
Thanks

Just to clear out your mind, try to see what is the value that outputs System.getProperty("user.dir") (let's call it A), this will print the complete absolute path from where your application was initialized, more specifically it will print the directory from where the JVM was started.
If you doesn't supply a parent path to the file that you are trying to open, the (A) path is taken by default and the file is looked inside that directory. So please, have in mind that.
Aditional information
If you absolutely need that file you should include it in your project so you can access it as a resource. Resource is a file that is included in your project and came bundled with the generated .jar or .war for re distribution.
My advice is to put the file in the package and use as a resource as it the safer way to work with external resources that should be shipped with your project.
Take a deeper look at this post for more about practical way of handling resources.

Please refer below link:
https://stackoverflow.com/a/2308388/1358551
You may have to use getResourceAsStream().

File path is working in NetBeans but not in built JAR file

I have a file holding default information that I use to load the textFields of my application. I looked up how to get this built into my jar file when I build and I was told to put it in the source packages and it would be brought along, so I have done that.
File Structure:
Project
-Source Packages
-src
~Java Classes
-defaultFiles
~Defaults.txt
The code I am trying to use is this:
BufferedReader in;
try {
URL resourceURL = FuelProperties.class.getResource("/defaultFiles/Defaults.txt");
in = new BufferedReader(new FileReader(resourceURL.getPath()));
}
And this works perfectly when I run it through NetBeans but when I build the project and try to run it from the jar file it is not grabbing the file.
I have verified that the default file is being built and exists in the same file structure shown above.
If you can help me out with this I would be extremely grateful as I have no idea what is keeping this from working. Thanks.

You have to lookup in the classpath, not on the disk.
The API to use is :
URL resourceURL : this.getClass().getResource("relative path in the classpath");
Once you have the url you can open a stream, etc.
EDIT : in the main method, you of course need to replace
this.getClass()
by
ClassName.class

I found the answer after searching through a couple dozen questions. It turns out that you can only get a InputStream of the data within a file within your JAR not a File object like I was attempting to do.
(If you want the File object you just have to extract the files from the JAR in your program and then you have access to it.)
So the code that got my problem to work was simply replacing this:
URL resourceURL = FuelProperties.class.getResource("/defaultFiles/Defaults.txt");
in = new BufferedReader(new FileReader(resourceURL.getPath()));
With this:
in = new BufferedReader(new InputStreamReader(this.getClass().getResourceAsStream("/defaultFiles/Defaults.txt")));
And now it is working both inside NetBeans and in the Built JAR file.

How selenium can test if it has read access to a file

Our test-app runs on multiple Virtual Machines through Selenium Remote Control.
The App sits on a test controller Server.
The test-app is used to test a third party online application.
How can I test to see if on certain VM Selenium-RC has read access to a file or folder.
Is there anything like file.canRead(filepath) kind of thing for selenium too?
Before you respond:
File's canRead(filepath) will only test if the file is readable from a test controller server, not able to say anything if it is readable on VM where actual browsers are opening(testing) third-party-online-application.
Basically, I want to upload some file to the third-party-online-application through selenium.
Before doing an upload, I want to make sure that the file is available for upload (on VMs).

A solution would be to create a download link in the application and then attempt to download the file via Selenium. That way, you get a user-representative experience.
If you want to be really fancy, have the Application create a file with the current date and then let the test download the file (simple text file) and check if the file contains the date. Then you test application writing a file and user reading the file, which covers access rights as well.

Which scripting language you are using? If assuming that your file to upload resides under "./data" directory then in java you can check with following steps
File file = new File("./data/myfile.ext");
boolean canUpload = file.exists() && file.canRead();
String fileToUpload = file.getCanonicalPath(); //file name with full path

File file = new File("Folder_Location"); // Folder path if file name not known
boolean canUpload = file.listFiles()[index].canRead();
Note : For latest downloaded file use
int size=file.listFiles().length-1;
boolean canUpload = file.listFiles()[size].canRead();