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How to read a file that i created inside my the same package?

This is a chunk of data I'd like to access by a method.
I'm doing the following to read my file:
String fileName = "file.txt"
InputStream inputStream = new FileInputStream(fileName);
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
My file.txt is in the same package, but I still get FileNotFoundException.
I didn't use a path url to point to the file because I thought since this it going to be an android application, hard-coding the path might not work when deployed... Please correct me if I am wrong. Thanks bunch!

This shows how to do that. https://stackoverflow.com/a/14377185/2801237
Also the 'package' your class is in has nothing to do with the 'path' where the file is being executed from. (two different concepts, 'package' = folder hierarchy of java source code files), 'path' = location on a filesystem of a specific file, your APK is being 'executed' in a particular place, and the location it writes a file is associated with that (I actually don't know where 'offhand' it writes by default, because I always get cache dir, or sd card root, etc.)

You may use:
InputStream inputStream = this.getClass().getResourceAsStream(fileName);

Programmatically uploading a file?

I'm developing a small program that uploads and downloads files from my box account.
I looked at the docs about uploading files and I found this code:
BoxFolder rootFolder = BoxFolder.getRootFolder(api);
FileInputStream stream = new FileInputStream("My File.txt");
rootFolder.uploadFile(stream, "My File.txt");
stream.close();
I don't really understand how it works. Where can I put the path to the file I want to upload? Or should I use different code?

The constructor for FileInputStream takes in a path to a file. The example in the documentation is uploading a file with the path "./My File.txt" relative to the current directory.
To make it a bit clearer, here's an example using a full path:
BoxFolder rootFolder = BoxFolder.getRootFolder(api);
FileInputStream stream = new FileInputStream("/path/to/My File.txt");
rootFolder.uploadFile(stream, "My File.txt");
stream.close();

Where i Can find text file created by servlet in Eclipse

This may be a stupid question, but I have to ask because I couldn't find any proper solution.
I am new to Eclipse. I created a Dynamic Web project in Eclipse, In this, I write a simple code to create a text file, Only file name is specified Not the path that where to create, After successful execution, i could not find my text file in my project folder.
If path is specified in the code, I can find the text file in specified directory, My Question is where i can find my text file if i am not specify a path ?
And my code is
try {
FileWriter outFile = new FileWriter("user_details.txt", true);
PrintWriter out1 = new PrintWriter(outFile);
out1.append(request.getParameter("un"));
out1.println();
out1.append(request.getParameter("pw"));
out1.close();
outFile.close();
System.out.println("file created");
} catch(Exception e) {
System.out.println("error in writing a file"+e);
}
I edited my code with following lines,
String path = new File("user_details.txt").getAbsolutePath();
System.out.println(path);
The path that i got is below
D:\Android\eclipse_JE\eclipse\user_details.txt
Why i got it in the eclipse folder ?
Then,
How can i create a text file in my web app, if this is not the right way to create a textfile ?

The file is located in the actual working directory of your application server. Do a
System.out.println(new File("").getAbsolutPath());
and you'll find the location.
However this is not a good idea to write files in web application like this, because first you never know where it is and second you never know whether you write privilege on it.
You need to specify some filesystem root for your application by passing it as init-parameter and use it as parent for everything you need to do on the filesystem. Check this answer to a similar Question.
You could then create your file like this:
String fsroot = getServletContext().getInitParameter("fsroot")
File ud = new File(fsroot, "user_details.txt");
FileWriter outFile = new FileWriter(ud, true);

You may try the getAbsolutePath() method.
String newFile = new File("Demo.txt").getAbsolutePath();
It will show the location where the files will be created.

Robovm Read and write to a file

I try to write a text to a file and read this text later. When I use FileWriter I become a NullPointerException?
Is that a permission problem or ...? I also try the PrintWriter but I see the same Exception
.
This my code:
FileWriter fw = new FileWriter(new File("file.file"));
fw.write("XYZ");
best regards
londi

I guess your problem is that you use a relative file path, but that the origin of the relative path is not the one you think.
First of all, try to use an absolute path, that would be, on linux-like machines something like /home/me/myCode/myfile.txt or on windows something like c:/some/path/myfile.txt
Another thing you can do, in order to know what happens is print the origin.
File origin = new File(".");
System.out.println(origin.getAbsolutePath());
Once you know where the origin is, you can see what you need in order to get to your file.
Hope it will help.

Sounds like a permission issue. On iOS your application lives within a security sandbox, so you cannot just randomly read and write files anywhere you want. You could either use File.createTempFile to create a temp file somewhere hidden you your sandbox where nothing else can see it, or use the native api to determine where to dump your files. The following example will give you a file reference to the Documents Directory folder:
NSArray nsa = NSFileManager.defaultManager().URLsForDirectory$inDomains$(NSSearchPathDirectory.DocumentDirectory, NSSearchPathDomainMask.UserDomainMask);
NSURL nsu = (NSURL)nsa.getFirst();
String snsu = nsu.getAbsoluteString() + "MyNewDocument.pdf";
File newFile = new File(new URI(snsu));

Load file dynamically from jar

I am trying to read a .json file I am packaging with my .jar.
The problem - finding the file so that I can parse it in.
The strange bit is that this code works in NetBeans, likely due to the way these methods work and the way NetBeans handles the dev workspace. When I build the jar and run it, however, it throws an ugly error: Exception in thread "main" java.lang.IllegalArgumentException: URI is not hierarchical.
My code for getting the file is as such:
//get json file
File jsonFile = new File(AndensMountain.class.getResource("/Anden.json").toURI());
FileReader jsonFileReader;
jsonFileReader = new FileReader(jsonFile);
//load json file
String json = "";
BufferedReader br = new BufferedReader(jsonFileReader);
while (br.ready()) {
json += br.readLine() + "\n";
}
I have gotten it to work if I allow it to read from the same directory as the jar, but this is not what I want - the .json is in the jar and I want to read it from in the jar.
I've looked around and as far as I can see this should work but it isn't.
If you are interested, this is the code before trying to get it to read out of the jar (which works as long as Anden.json is in the same directory as AndensMountain.jar):
//get json file
String path = AndensMountain.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath();
File jsonFileBuilt = new File(new File(path).getParentFile(), "Anden.json");
File jsonFileDev = new File(new File(path), "Anden.json");
FileReader jsonFileReader;
try {
jsonFileReader = new FileReader(jsonFileBuilt);
} catch (FileNotFoundException e) {
jsonFileReader = new FileReader(jsonFileDev);
}

Try
Reader reader = new InputStreamReader(AndensMountain.class.getResourceAsStream("/Anden.json"), "UTF-8");
AndensMountain.class.getResource("/Anden.json") URL when ran outside a jar (for example, when the classes are compiled to a "classes/" directory) is a "file://" URL.
That is not the case when ran from inside a jar: it then becomes a "jar://" URL.
The java.io.File doesn't know how to handle this type of URL. It handles only "file://".
Anyway you don't really need to treat it as a File. You can manipulate the URL itself (either to navigate to a parent directory, for example) or to get its contents (via openStream(), or if you need to add headers, via openConnection()).
java.lang.Class#getResourceAsStream() as I suggested is just shorthand to Class#getResource() followed by openStream() on its result.