Develop Reference

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

What type of File is created when we create new File without the extension?

In java, when we create a file, we create files using the name of the extension.
For example :
File file = new File("D:/light.txt");
I would like to know what type of file format do we get when we create a file without the file extension type.
For example :
File file = new File("D:/light");

This answer assumes you're doing more than just creating a File object - that you're actually creating a file on the file system. (A File object is just a logically representation of a file system entry which may or may not exist.) If you're really just creating a File object, read EJP's answer - at that point, you've basically just got a name. That doesn't have a "type" or a "format".
The extension is just part of the name. The operating system may try to use that to display a different icon, or launch a specific application when you double-click on the icon, or whatever - but it's really just part of the name.
Fundamentally, a file consists of:
The name you specify when you create it
The bytes you write in it
Metadata such as access control
Unless you deliberately add metadata, it's typically just inherited (default permissions etc).
You can write any data in any file - just because a file has an extension of .txt doesn't mean it's definitely a text file. It could have content which is actually MP3-encoded audio data, for example. Whether the OS uses the file extension or the content to work out what to do with the file is up to the OS.

What type of File is created when we create new File without the extension?
No file is created at all, and there is only one type of File.
In java, when we create a file, we create files using the name of the extension.
Or not.
For example: File file = new File("D:/light.txt");
I would like to know what type of file format do we get when we create a file without the file extension type.
You don't. You don't get any file format at all, because you don't get a file: only a File object in memory.
For example: File file = new File("D:/light");
You can produce all the examples you want, but no file is created, and no file format.
In any case Java doesn't care about filename extensions. Your operating system might, but that's a different story.

Extracting files from res folder in Executable JAR (txt files to be specific)

I would like to ask if its possible to put text files into my jar, I use them to make my map in my game, but users can get Highscores. now I want to save the Highscores with the map, so I have to save the map on the user their PC. Is there any way how I could do this? I've searched the internet for some ideas but I could not find anything that even came close to what I've wanted. I only had 3/4th of a year java so I don't know much about these things, everything that happens outside the debug of eclipse are problems for me(files are mainly one of those things, null exceptions, etc).
The main question now.
Is it possible to do? If yes, do you have any terms I could search on, or some sites/guides/tutorials? If no, is there any other way how I could save the highscores?
EDIT:
to make clear
Can I get the text file (the text inside the file) to be extracted to a different file in like the home directory of my game (where I save the settings and stuff) the basic maps are inside the jar file, so I want them to be extracted on the first start-up of the program
Greetings Carolien

"extracted to a different file in like the home directory of my game (where i save the settings and stuff) the basic maps are inside the jar file, so i want them to be extracted on the first startup of the program"
You can get the URL by using getClass().getResource()
URL url = getClass().getResource("/res/myfile.txt");
Then create a File object from the URI of the URL
File file = new File(url.toURI());
Then just perform your normal file operations.
if (file.renameTo(new File(System.getProperty("user.home") + "\\" + file.getName()))) {
System.out.println("File is moved successful!");
} else {
System.out.println("File is failed to move!");
}
Assuming your file structure is like below, it should work fine
ProjectRoot
src
res
myfile.txt
Note: the above is moving the entire file. If you want to extract just the data inside the file, then you can simple use
InputStream is = getClass().getResourceAsStream("/res/myfile.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
The just do normal IO operation with the reader. See here for help with writing the file.

Can I use File.createTempFile() to create a file with a non-random name

I want to create a temporary file (that goes away when the application closes) with a specific name. I'm using this code:
f = File.createTempFile("tmp", ".txt", new File("D:/"));
This creates something like D:\tmp4501156806082176909.txt. I want just D:\tmp.txt. How can I do this?

In this case, don't use createTempFile. The point of createTempFile is to generate the "garbage" name in order to avoid name colisions.
You should use File.createNewFile() or simply write to the file. Whichever is more appropriate for your use case. You can then call File.deleteOnExit() to get the VM to look after cleaning up the file.

If you want to create just tmp.txt, then just create the file using createNewFile(), instead of createTempFile(). createTempFile is used to create temporary files that should not have the same name when created over and over.
Also have a look at this post which shows a very simple way to create files.
Taken the post mentioned above:
String path = "C:"+File.separator+"hello"+File.separator+"hi.txt";
//(use relative path for Unix systems)
File f = new File(path);
//(works for both Windows and Linux)
f.mkdirs();
f.createNewFile();

try regex
fileName = fileName.replaceAll("\\d", "");

Save my Unique ID (String) and retrieving it later when I re-launch my Java application

I want to save a Unique ID (which is a String) which gets created when I launch my Java application. Now I want to save this somewhere (I think in some file on the disk) so that when I relaunch my application I should be able to read it and use that ID.
I want to know what is the good way to saving such ID. I am thinking of creating a Properties file and save it then retrieve it from it when I relaunch application. Is there a better or standard way for this?
EDIT :
Additionally what should be the folder location for storing on the disk. Should it be relative to my execution path or some Logged-in user specific path?

1. If its the same Java application that writes or reads this String, then use Serialization, it will be in non-readable form when saved.
2. If reading and writing is from different program, then use Text file.
3. Using Property file will be also a good approach.

If your app/program needs to store more data at some point sqlite3 might be the best option for you. It is easy to implement and use.
Download sqlite3
EDIT: How many IDs will be stored in the app? If there are just a few, a textfile or property file is enough.
EDIT2: Navigate to your Documents folder on your machine and you will see folders of programs/games. Thats where you should place the file/db. However you can also store it in the installation path on your hard drive. Also make sure your user launches the app trough a shortcut, not the actual execution file

Use the FileWriter and File classes from Java.
It should be something like that:
File f = new File(your path here);
if (f.exists()){
BufferedReader br = new BufferedReader(new FileReader(your path here));
String a = br.readLine();
br.close();
}else{
FileWriter fw = new FileWriter(your path here);
fw.write(your ID String);
fw.flush();
fw.close();
I hope this is want u meant.
Best regards
edit: just noticed too late that your edited your post....