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How to read a file that is located in any location on the computer?

So I have a project, and this is one of the demands:
You should have a class named Project3, containing a main method.
This program reads the levels information from a file whose name is
specified as a command-line parameter (The file should also be
relative to the class-path as described here:)
All the file names specified in the levels and block definition files
should be relative to the class path. The reason we want them to be
relative to the class path is that later we will be able to read the
files from inside a jar, something we can not do with regular File
references.
To get an input stream relative to the class path (even if it's inside
a jar), use the following:
InputStream is =
ClassLoader.getSystemClassLoader().getResourceAsStream("image.png");
The idea is to keep a folder with files(definitions and images) and
then add that folder to the class path when running the JVM:
java -cp bin:resources ... If you don't add the resources folder to
you class path you wont be able to load them with the command from
above.
When run without parameters, your program should read a default level
file, and run the game accordingly. The location of the default level
file should be hard-coded in your code, and be relative to the
classpath_.
When run without parameters, your program should read a default level file, and run the game accordingly. The location of the default level file should be hard-coded in your code, and be relative to the classpath_.
The part of the code that handles the input is:
public Void run() throws IOException {
LevelReader level = new LevelReader();
List<level> chosenLevels = new ArrayList<>();
if (args.length >= 1) {
File f = new File(args[0]);
if (f.exists()) {
chosenLevels = level.makeLevel(args[0]);
}
}
if (chosenLevels.size() == 0) {
game.runLevels(defaultLevels);
} else {
game.runLevels(chosenLevels);
}
return null;
}
So my question is:
An argument should be the full path of a file which means:
D:\desktop\level3.txt
Is it possible to read a file from every location on my computer?
Because right now I can do it only if my text file is in the
project's directory (not even in the src folder).
I can't understand the rest of their demands. What does is mean "should be hard-coded in your code, and be relative to the
classpath_." and why is it related to InputStream method(?)
I'm confused all over this.
Thanks.

A classpath resource is not the same as a file.
As you have correctly stated, the full path of a file is something like D:\desktop\level3.txt.
But if ever want to distribute your application so it can run on other computers, which probably won’t have that file in that location, you have two choices:
Ask the user to tell the program where to find the file on their computer.
Bundle the file with the compiled program.
If you place a non-.class file in the same place as .class files, it’s considered a resource. Since you don’t know at runtime where your program’s class files are located,¹ you use the getResource or getResourceAsStream method, which is specifically designed to look in the classpath.
The getResource* methods have the additional benefit that they will work both when you are developing, and when the program is packaged as a .jar file. Individual entries in a .jar file are not separate files and cannot be read using the File or FileInputStream classes.
If I understand your assignment correctly, the default level file should be an application resource, and the name of that resource is what should be hard-coded in your program. Something like:
InputStream is;
if (args.length > 0) {
is = new BufferedInputStream(
new FileInputStream(args[0]));
} else {
// No argument provided, so use program's default level data.
is = ClassLoader.getSystemClassLoader().getResourceAsStream("defaultlevel.txt");
}
chosenLevels = level.makeLevel(is);
¹ You may find some pages that claim you can determine the location of a running program’s code using getProtectionDomain().getCodeSource(), but getCodeSource() may return null, depending on the JVM and ClassLoader implementation, so this is not reliable.

To answer your first question, it doesn't seem like they're asking you to read from anywhere on disk, just from within your class path. So that seems fine.
The second question, "What does is mean 'should be hard-coded in your code, and be relative to the classpath'?". You are always going to have a default level file in your project directory. Define the path to this file as a String in your program and that requirement will be satisfied. It's related to the InputStream because the stream requires a location to read in from.

Get current path of executed file

I try to write and read to the file in my java project file called Books.txt.
The problem is that I can access the file only if partialPath has full path to the file.
Here is the code:
public <T> List<T> readFromFile(String fileName) {
private String partialPath = "\\HW3\\src\\java\\repos\\";
try {
String path = partialPath + fileName;
FileInputStream fi = new FileInputStream(path);
ObjectInputStream oi = new ObjectInputStream(fi);
// Read objects
List<T> items = (List<T>) oi.readObject();
oi.close();
fi.close();
return items;
} catch (IOException | ClassNotFoundException e) {
}
}
If I set relative path as above I get exception file not found.
My question is how can I set full path to the current directory programmatically?

Here is a code snippet of the Drombler Commons - Client Startup code I wrote, to determine the location of the executable jar. Replace DromblerClientStarter with your main class.
This should work at least when you're running your application as an executable JAR file.
/**
* The jar URI prefix "jar:"
*/
private static final String FULL_JAR_URI_PREFIX = "jar:";
/**
* Length of the jar URI prefix "jar:"
*/
private static final int FULL_JAR_URI_PREFIX_LENGTH = 4;
private Path determineMainJarPath() throws URISyntaxException {
Class<DromblerClientStarter> type = DromblerClientStarter.class;
String jarResourceURIString = type.getResource("/" + type.getName().replace(".", "/") + ".class").toURI().
toString();
int endOfJarPathIndex = jarResourceURIString.indexOf("!/");
String mainJarURIString = endOfJarPathIndex >= 0 ? jarResourceURIString.substring(0, endOfJarPathIndex)
: jarResourceURIString;
if (mainJarURIString.startsWith(FULL_JAR_URI_PREFIX)) {
mainJarURIString = mainJarURIString.substring(FULL_JAR_URI_PREFIX_LENGTH);
}
Path mainJarPath = Paths.get(URI.create(mainJarURIString));
return mainJarPath;
}
Depending on where you bundle Books.txt in your application distribution package, you can use this mainJarPath to determine the path of Books.txt.

I also feel that files created (and later possibly modified and or deleted) by your running Java application is usually better to be placed in a location of the file system that is away from your java application installed home directory. An example might be the 'C:\ProgramData\ApplicationNameFiles\' for the Windows operating system or something similar for other OS platforms. In my opinion, at least for me, I feel it provides less chance of corruption to essential application files due to a poorly maintained drive or, accidental deletion by a User that opens up a File Explorer and decides to take it upon him/her self to clean their system of so called unnecessary files, and other not so obvious reasons.
Because Java can run on almost any platform and such data file locations are platform specific the User should be allowed to select the location to where these files can be created and manipulated from. This location then can be saved as a Property. Indeed, slightly more work but IMHO I feel it may be well worth it.
It is obviously much easier to create a directory (folder) within the install home directory of your JAR file when it's first started and then store and manipulate your application's created data files from there. Definitely much easier to find but then again...that would be a matter of opinion and it wouldn't be mine. Never-the-less if you're bent on doing it this way then your Java application's Install Utility should definitely know where that install path would be, it is therefore just a matter of storing that location somewhere.
No Install Utility? Well then your Java application will definitely need a means to know from where your JAR file is running from and the following code is one way to do that:
public String applicationPath(Class mainStartupClassName) {
try {
String path = mainStartupClassName.getProtectionDomain().getCodeSource().getLocation().getPath();
String pathDecoded = URLDecoder.decode(path, "UTF-8");
pathDecoded = pathDecoded.trim().replace("/", File.separator);
if (pathDecoded.startsWith(File.separator)) {
pathDecoded = pathDecoded.substring(1);
}
return pathDecoded;
}
catch (UnsupportedEncodingException ex) {
Logger.getLogger("applicationPath() Method").log(Level.SEVERE, null, ex);
}
return null;
}
And here is how you would use this method:
String appPath = applicationPath(MyMainStartupClassName.class);
Do keep in mind that if this method is run from within your IDE it will most likely not return the path to your JAR file but instead point to a folder where your classes are stored for the application build.

This is not a unique issue to Java, it's a problem faced by any developer of any language wishing to write data locally to the disk. The are many parts to this problem.
If you want to be able to write to the file (and presumably, read the changes), then you need to devise a solution which allows you find the file in a platform independent way.
Some of the issues
The installation location of the program
While most OS's do have some conventions governing this, this doesn't mean they are always used, for what ever reason.
Also, on some OS's, you are actively restricted from writing to the "installation" location. Windows 8+ doesn't allow you to write to the "Program Files" directory, and in Java, this usually (or at least when I was dealing with it) fails silently.
On MacOS, if you're using a "app bundle", the working directory is automatically set to the user's home directory, making it even more difficult to manage
The execution context (or working directory) may be different from the installation location of the program
A program can be installed in one location, but executed from a different location, this will change the working directory location. Many command line tools suffer from this issue and use different conventions to work around it (ever wonder what the JAVA_HOME environment variable is for 🤔)
Restricted disk access
Many OS's are now actively locking down the locations to which programs can write, even with admin privileges.
A reusable solution...
Most OS's have come up with conventions for solving this issue, not just for Java, but for all developers wishing to work on the platform.
Important Like all guide lines, these are not hard and fast rules, but a recommendations made by the platform authors, which are intended to make your life simpler and make the operation of the platform safer
The most common solution is to simply place the file in a "well known location" on the disk, which can be accessed through an absolute path independently of the installation or execution location of the program.
On Windows, this means placing the file in either ~\AppData\Local\{application name} or ~\AppData\Roaming\{application name}
On MacOS, this means placing the file in ~/Library/Application Data/{application name}
On *nix, this typically means placing the file in ~/.{application name}
It could be argued that you could use ~/.{application name} on all three platforms, but as a user who "shows hidden files", I'd prefer you didn't pollute my home directory.
A possible, reusable, solution...
When Windows 8 came out, I hit the "you can't write to the Program Files" issue, which took some time to diagnose, as it didn't generate an exception, it just failed.
I was also working a lot more on Mac OS as well, so I needed a simple, cross platform solution, so my code could automatically adapt without the need for multiple branches per platform.
To this end, I came with a simple utility class...
public enum SystemUtilities {
INSTANCE;
public boolean isMacOS() {
return getOSName().startsWith("Mac");
}
public boolean isMacOSX() {
return getOSName().startsWith("Mac OS X");
}
public boolean isWindowsOS() {
return getOSName().startsWith("Windows");
}
public boolean isLinux() {
return getOSName().startsWith("Linux");
}
public String getOSName() {
return System.getProperty("os.name");
}
public File getRoamingApplicationSupportPath() {
// For *inx, use '~/.{AppName}'
String path = System.getProperty("user.home");
if (isWindowsOS()) {
path += "\\AppData\\Roaming";
} else if (isMacOS()) {
path += "/Library/Application Support";
}
return new File(path);
}
public File getLocalApplicationSupportPath() {
// For *inx, use '~/.{AppName}'
String path = System.getProperty("user.home");
if (isWindowsOS()) {
path += "\\AppData\\Local";
} else if (isMacOS()) {
path += "/Library/Application Support";
}
return new File(path);
}
}
This provides a baseline from which "independent" code can be built, for example, you could use something like...
File appDataDir = new File(SystemUtilities.INSTANCE.getLocalApplicationSupportPath(), "MyAwesomeApp");
if (appDataDir.exists() || appDataDir.mkdirs()) {
File fileToWrite = new File(appDataDir, "Books.txt");
//...
}
to read/write to the file. Although, personally, I might have manager/factory do this work and return the reference to the end File, but that's me.
What about "pre-packaged" files?
Three possible solutions...
Create the file(s) if they don't exist, populating them with default values as required
Copy "template" file(s) out of the Jar file, if they don't exist
Use an installer to install the files - this is the solution we used when we were faced with changing the location of all our "external" configuration files.
Read only files...
For read only files, the simplest solution is to embedded them within the Jar as "embedded resources", this makes it easier to locate and manage...
URL url = getClass().getResource("/path/to/readOnlyResource.txt");
How you do this, will depend on your build system

Java load external CSV file not working after compiling to Jar

I am writing a Java program using NetBeans that times the process of developing black and white film. One of the options allows you to set the times manually and the other allows you to select the chosen film and developer to calculate the required times from a CSV file that is loaded into an array.
There are 3 files that are loaded, filmdb.csv, masterdb.csv and usersettings.csv. The first two are loaded purely for reading and the third file is loaded and can be written too to save the users default settings.
All 3 files are stored in the project directory and loaded in a similar way to the following code and called from main:
static String[] filmArray;
static int filmRows = 125;
int selectedDevTime;
int tempDevTime;
int minDevTime;
int secDevTime;
public int count = -1;
static void createFilmArray() {
filmArray = new String[filmRows];
Scanner scanLn = null;
int Rowc = 0;
int Row = 0;
String InputLine = "";
String filmFileName;
filmFileName = "filmdb.csv";
System.out.println("\n****** Setup Film Array ******");
try {
scanLn = new Scanner(new BufferedReader(new FileReader(filmFileName)));
while (scanLn.hasNextLine()) {
InputLine = scanLn.nextLine();
filmArray [Rowc] = InputLine;
Rowc++;
}
} catch (Exception e) {
System.out.println(e);
}
}
When I press the run button in NetBeans all works well, the files are all loaded and stored to the appropriate array but when I build the file as a .jar the files are not loaded, I have tried copying the 3 files to the same directory as the .jar as well as importing the 3 files into the .jar archive but to no joy.
Is there a specific location where the files should be placed, or should they be imported in a different way?
I would rather not load them from an absolute directory as for example a Windows user may have the files stored in C:\users\somebody\devtimer\file.csv and a Linux user may have the files stored in /home/somebody/devtimer/file.csv.
Cheers
Jim

your files got packaged into the *.jar
once a file is in the jar you cannot accessit as a file using a path (since, if you think about it, its no longer a file. its inside another file, the jar file).
you need to do something like this:
InsputStream csvStream = this.getClass().getResourceAsStream("filmdb.csv");
then you can pass this input stream to your csv parser.
you will not be able to make any modifications to the fhile though. any file you want to change you will need to store outside of your jar.
as for file paths, things like new File("somename") are resolved relative to the current working directory. if you want to know where your root is try something like:
System.err.println(new File(".").getCanonicalPath());
this will be very sensitive to what the working directory was when your application was executed, which in turn depends on how exactly it was executed etc.

If you are running the jar from the same directory the csv files are in then it should be loading. There might be some issue though if you just double click on the jar. You could create a new File represented by the filename and check the absolute path to find out where the jar is running from.
If you want to include the csv inside the jar (cleaner for distribution), you need to load the files in a different way. See this question for a nice example.

Unable to open resources in directories which end with an exclamation mark (!)

The getResourceAsStream-method returns null whenever running the executable jar in a directory which ends with a exclamation mark.
For the following example, I have a Eclipse project the following directory structure:
src\ (Source Folder)
main\ (Package)
Main.java
res\ (Source Folder)
images\
Logo.png
I'm reading the Logo.png as follows:
public static void main(String[] args) throws IOException {
try (InputStream is = Main.class.getClassLoader().getResourceAsStream("images/Logo.png")) {
Image image = ImageIO.read(is);
System.out.println(image);
}
}
See the attachment for 2 test cases. First, the executable jar is started from the directory "D:\test123!##" without any problems. Secondly, the executable jar is started from the directory "D:\test123!##!!!", with problems.
Are directories ending with an exclamation mark not supported? Is the code wrong?
Thanks in advance.

Probably because of this bug or any of the many similar bugs in the Java bug database:
http://bugs.sun.com/view_bug.do?bug_id=4523159
The reason is that "!/" in a jar URL is interpreted as the separator between the JAR file name and the path within the JAR itself. If a directory name ends with !, the "!/" character sequence at the end of the directory is incorrectly interpreted. In your case, you are actually trying to access a resource with the following URL:
jar:file:///d:/test1231##!!!/test.jar!/images/Logo.png
The bug has been open for almost 12 years and is not likely to be fixed. Actually I don't know how it can be fixed without breaking other things. The problem is the design decision to use ! as a character with a special meaning (separator) in the URL scheme for JAR files:
jar:<URL for JAR file>!/<path within the JAR file>
Since the exclamation mark is an allowed character in URLs, it may occur both in the URL to the JAR file itself, as well as in the path within the JAR file, making it impossible in some cases to find the actual "!/" separator.

A simple work around for Windows is to use "\" instead of "/" in the path. That would mean the "!/" character sequence is found after the full path. For instance:
new URL("jar:file:\\d:\\test1231##!!!\\test.jar!/images/Logo.png");
My Code:
File jar = new File(jarPath + "/" + jarName);
URL url = new URL("jar:" + jar.toURI() + "!" + dataFilePath);
InputStream stream = null;
try {
stream = url.openStream();
} catch (FileNotFoundException e) {
// Windows fix
URL urlFix = new URL("jar:" + jar.toURI().toString().replace('/', '\\')
+ "!" + dataFilePath);
stream = urlFix.openStream();
}
I use toURI() because it handles things like spaces.
Fixes:
The fix itself would be for Java to check if the file exists and if not continue to the next separator (the "!/" part of the url) until the separators are exhausted, then throw the exception. So it would see that "d:\test1231##!!" throws a java.io.FileNotFoundException and would then try "d:\test1231##!!!\test.jar" which does exist. This way it does not matter if there are "!" in the file path or in the jar's files.
Alternatively the "!/" can be switched to something else that is an illegal file name or to something specific (like "jarpath:").
Alternatively make the jar's file path use another parameter.
Note:
It may be possible to override something, swap a handler, or change the code to open the file first then look inside the jar file later but I have not looked.

How to read from a file that has no extension in Java?

So basically say i have a file that is simply called settings, however it has no extension, but contains the data of a text file renamed.
How can i load this into the file() method in java?
simply using the directory and file seems to make java think its just a directory and not a file.
Thanks

In Java, and on unix, and even on the filesystem level on windows, there is no difference in if a file has an extension or not.
Just the Windows Explorer, and maybe its pendants on Linux, use the extension to show an appropriate icon for the file, and to choose the application to start the file with, if it is selected with a double click or in similar ways.
In the filesystem there are only typed nodes, and there can be file nodes like "peter" and "peter.txt", and there can be folder nodes named "peter" and "peter.txt".
So, to conclude, in Java there is really no difference in file handling regarding the extension.

new File("settings") should work fine. Java does not treat files with or without extension differently.

Java doesn't understand file extensions and doesn't treat a file any differently based on its extension, or lack of extension. If Java thinks a File is a directory, then it is a directory. I suspect this is not what is happening. Can you try?
File file = new File(filename);
System.out.println('\'' + filename + "'.isDirectory() is "+file.isDirectory());
System.out.println('\'' +filename + "'.isFile() is "+file.isFile());
BTW: On Unix, a file file. is different to file which is different to FILE. AFAIK on Windows/MS-DOS they are treated as the same.

The extension should not make a difference. Can you post us the code you are using? And the error message please (stack trace).
Something along these lines should do the trick (taken from http://www.kodejava.org/examples/241.html)
//
// Create an instance of File for data file.
//
File file = new File("data");
try {
//
// Create a new Scanner object which will read the data
// from the file passed in. To check if there are more
// line to read from it we check by calling the
// scanner.hasNextLine() method. We then read line one
// by one till all line is read.
//
Scanner scanner = new Scanner(file);
while (scanner.hasNextLine()) {
String line = scanner.nextLine();
System.out.println(line);
}
} catch (FileNotFoundException e) {
e.printStackTrace();
}