Java input == null why?

Question

I'm using a simple way to get my resources for the project. I'm using Eclipse, and I have a 'res' folder to hold the needed files. This is how I load stuff, for example a 'puppy.png' just in my res folder (no subfolders):

String path = "/puppy.png";
try {
    BufferedImage image = ImageIO.read(getClass().getResourceAsStream(path));
} catch(Exception ex) { ex.printStackTrace(); }

And sometimes I get an input==null error, and sometiomes not! Not like this time puppy.png loaded but next time it won't. For some classes it always loads correctly, and for the other classes I always get this error. Can anyone explain why can this happen, and how can I fix it, but still use the getResourceAsStream() method?

Try using a backslash instead of a forward slash. It may or may not help. EDIT: You may not even need the slash at all — DerStrom8
– DerStrom8, Commented Apr 22, 2014 at 19:32
That's why I provided it as a comment rather than an answer. And see my edit, I realized the possibility that I was wrong anyway — DerStrom8
– DerStrom8, Commented Apr 22, 2014 at 19:35
Maybe you should also print out the classloader of the class... — keuleJ
– keuleJ, Commented Apr 22, 2014 at 19:40

Community · Accepted Answer · 2017-05-23 11:49:24Z

1

Please have a look at How to retrieve image from project folder?.

I have mentioned no of ways to read image from different paths.

You can try any one

// Read from same package 
ImageIO.read(getClass().getResourceAsStream("c.png"));

// Read from absolute path
ImageIO.read(new File("E:\\SOFTWARE\\TrainPIS\\res\\drawable\\c.png"));

// Read from images folder parallel to src in your project
ImageIO.read(new File("images\\c.jpg"));

In your case the image must be in the same package where is the class and don't prefix /.

edited May 23, 2017 at 11:49

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answered Apr 22, 2014 at 19:36

Braj

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1 Comment

keuleJ Over a year ago

Yes also see the javadoc of the method: Before delegation, an absolute resource name is constructed from the given resource name using this algorithm: If the name begins with a '/' ('\u002f'), then the absolute name of the resource is the portion of the name following the '/'. Otherwise, the absolute name is of the following form: modified_package_name/name Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e').

Rogue · Accepted Answer · 2014-04-22 19:39:02Z

0

Note that if the resource returns null (meaning it doesn't exist), you will get this error.

Check the input returned like so:

String path = "/puppy.png";
try {
    InputStream is = getClass().getResourceAsStream(path);
    if (is == null) {
        //resource doesn't exist
    } else {
        BufferedImage image = ImageIO.read(is);
    }
} catch(Exception ex) { ex.printStackTrace(); }

Note that you most likely should be using String path = "puppy.png", seeing as you will already be in the content of the project folder.

edited Apr 22, 2014 at 19:39

answered Apr 22, 2014 at 19:33

Rogue

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7 Comments

LPeter1997 Over a year ago

How could it be? it does exist!

Rogue Over a year ago

If it does not find it, then it will return in that form. Try checking the value returned from getClass().getResourceAsStream(path)

Sotirios Delimanolis Over a year ago

Careful. getResource does not expect a file system specific separator. It expects / literally.

Rogue Over a year ago

Aha, in that case I'll change my answer then.

Luiggi Mendoza Over a year ago

The main problem is that the resource exists but sometimes cannot be found. This would treat it safely but doesn't solve the problem.

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