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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.

File path issue on windows vs Unix in java

In my program, I am reading a resource file for a unit test. I use file path as:
\\\path\\\to\\\file
On my machine(Windows) this runs fine. But on server(Unix), this fails, and I have to change it to: /path/to/file
But Java is supposed to be platform independent. So isn't this behaviour unexpected?

Use FileSystem.getSeparator() or System.getProperty("file.separator") instead of using slashes.
EDIT:
You can get an instance of FileSystem via FileSystems.getDefault (JDK 1.7+)

You can use File.separator to get the appropriate character in a platform-independent way.

Java is platform independent. The file path-es and some system calls are not.
As long as the path is relative, you can use File.separator:
String path = "path" + File.separator + "to" + File.separator + "file";
System.out.println(path); // prints path\to\file on windows
Sometimes it's an option is to provide a Properties file and let the user define path of that actual file. This way full paths are okay too. You can read the properties like this:
Properties props = new Properties();
props.load(new FileInputStream(filePath));
The next question is: how to specify the location of that file? That might be either a file on a relative path. If that's not viable for your app, then you can let the user specify it in a system property:
java ... -DconfigFile=C:\TEMP\asd.txt .... -jar myapp.jar
Then you can access it like this:
// prints C:\TEMP\asd.txt if you specified -DconfigFile=C:\TEMP\asd.txt
System.out.println(System.getProperty("configFile"));

This is the expected behaviour.
Java code compiles on any machine/OS provided you have the right version of Java installed on it.
However, at run time, your code sees only a variable value like another one, which happens to be \path\to\file
When it talks to the file system, it uses that particular value ; the file system then tries to find that path you've given to it ; which is why one syntax works fine on Windows but will not work on Linux.

Better way of doing this is :
val pathUri = Paths.get(".//src//test//res//file.txt").toUri()
val is = FileInputStream((File(pathUri)))

Storing Instances with Serialization

I am making a program that needs to save objects for retrieval at a future date. The program will be given out away as a jar file to different people.
I can already store and retrieve instances of classes when giving the Object input/output stream a absolute path (String) as a parameter.
I can also save images and text files in the resources folder and get it as a resource with getClass().getResource(String path).
Here is the problem:
I have tried every way possible to save/get Objects to/from the resources folder. It gets really weird dealing with URLS and Files and not ordinary Strings. Can someone please help me? I need to be able to save and retrieve objects relative to the classpath so that i can access the objects when the program is a jar file saved in different paths on the computer.

1: resource folder (in jar), is read-only.
You can create datas, store in the jar when you package, but after, it is finished: only to read.
2: so you want user can read and write (and it is not embedded in your app).
if it is personal datas, you can use (for PC):
String appdata= System.getenv("APPDATA");
System.out.println(appdata);
String dataFolder = System.getProperty("user.home") + "\\Local Settings\\ApplicationData";
System.out.println(dataFolder);
String dataFolder2 = System.getenv("LOCALAPPDATA");
System.out.println(dataFolder2);
on my PC, it gives:
C:\Users\develop2\AppData\Roaming
C:\Users\develop2\Local Settings\ApplicationData
C:\Users\develop2\AppData\Local
see this: What is the cross-platform way of obtaining the path to the local application data directory?
it is is for everybody, same principles, but you can encounter security issues
like this:
String programdata = System.getenv("PROGRAMDATA");
System.out.println(programdata);
String allusersprofile = System.getenv("ALLUSERSPROFILE");
System.out.println(allusersprofile); // same thing !
String publicdir = System.getenv("PUBLIC");
System.out.println(publicdir);

Is it possible to read a shapefile using geotools WITHOUT specifying the url of the file?

I am creating a web application which will allow the upload of shape files for use later on in the program. I want to be able to read an uploaded shapefile into memory and extract some information from it without doing any explicit writing to the disk. The framework I am using (play-framework) automatically writes a temporary file to the disk when a file is uploaded, but it nicely handles the creation and deletion of said file for me. This file does not have any extension, however, so the traditional means of reading a shapefile via Geotools, like this
public void readInShpAndDoStuff(File the_upload){
Map<String, Serializable> map = new HashMap<>();
map.put( "url", the_upload.toURI().toURL() );
DataStore dataStore = DataStoreFinder.getDataStore( map );
}
fails with an exception which states
NAME_OF_TMP_FILE_HERE is not one of the files types that is known to be associated with a shapefile
After looking at the source of Geotools I see that the file type is checked by looking at the file extension, and since this is a tmp file it has none. (Running file FILENAME shows that the OS recognizes this file as a shapefile).
So at long last my question is, is there a way to read in the shapefile without specifying the Url? Some function or constructor which takes a File object as the argument and doesn't rely on a path? Or is it too much trouble and I should just save a copy on the disk? The latter option is not preferable, since this will likely be operating on a VM server at some point and I don't want to deal with file system specific stuff.
Thanks in advance for any help!

I can't see how this is going to work for you, a shapefile (despite it's name) is a group of 3 (or more) files which share a basename and have extensions of .shp, .dbf, .sbx (and usually .prj, .sbn, .fix, .qix etc).
Is there someway to make play write the extensions with the tempfile name?

How to Move file in the same storage by changing path sytem level Android

How to move file not copy by just changing the path of file system level in android I have path like this
File f = new File(/storage/Folder1/Folder2/image.png);
File newfile = new File((/storage/Folder3/image.png);
I want to change the path of f to newfile without coping because it takes time and system give us support if we are in same mount point we can move file super fast just like if we move file in dextop windows in the same drive then speed is so fast I want to achieve same thing
Please give some sample code.

You can use Files.move, with options for retaining file attributes and detailed error reporting via several exceptions:
try {
Files.move( f.toPath(), newFile.toPath() );
} catch(...){
...
}
Possibly also the simpler method works for you, although this is more implementation dependent:
f.rename( newFile );