Hello I have this in my code
File file = new File("words.txt");
Scanner scanFile = new Scanner(new FileReader(file));
ArrayList<String> words = new ArrayList<String>();
String theWord;
while (scanFile.hasNext()){
theWord = scanFile.next();
words.add(theWord);
}
But for some reason I am getting a
java.io.FileNotFoundException
I have the words.txt file in the same folder as all of my .java files
What am I doing wrong? Thanks!
Tip: add this line to your code...
System.out.println(file.getAbsolutePath());
Then compare that path with where your file actually is. The problem should be immediately obvious.
The file should reside in the directory from which you execute the application, i.e. the working directory.
Generally it's a good idea to package data files with your code, but then using java.io.File to read them is a problem, as it's hard to find them. The solution is to use the getResource() methods in java.lang.ClassLoader to open a stream to a file. That way the ClassLoader looks for your files in the location where your code is stored, wherever that may be.
try:
URL url = this.getClass().getResource( "words.txt" );
File file = new File(url.getPath());
InputStream stream = this.getClass().getResourceAsStream( "words.txt" );You haven't specified an absolute path. The path would therefore be treated as a path, relative to the current working directory of the process. Usually this is the directory from where you've launched the Main-Class.
If you're unsure about the location of the working directory, you can print it out using the following snippet:
System.out.println(System.getProperty("user.dir"));
Fixing the problem will require adding the necessary directories in the original path, to locate the file.
java.io.FileReader.FileReader( java.io.File )states that aFileNotFoundExceptionis thrown "if the file does not exist, is a directory rather than a regular file, or for some other reason cannot be opened for reading." If you're sure the file exists and that it's a regular file, then there must be some reason it can't be opened. Make sure the file's permissions allow for reading as well.