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, Apr 22 '14 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, Apr 22 '14 at 19:35
Maybe you should also print out the classloader of the class... — keuleJ, Apr 22 '14 at 19:40

score 1 · Accepted Answer · edited May 23 '17 at 11:49

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 '17 at 11:49

Community

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answered Apr 22 '14 at 19:36

Braj

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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'). – keuleJ Apr 22 '14 at 19:41

Rogue · Answer 2 · 2014-04-22T19:39:02.357

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 '14 at 19:39

answered Apr 22 '14 at 19:33

Rogue

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How could it be? it does exist! – LPeter1997 Apr 22 '14 at 19:34
If it does not find it, then it will return in that form. Try checking the value returned from `getClass().getResourceAsStream(path)` – Rogue Apr 22 '14 at 19:35
1

Careful. `getResource` does not expect a file system specific separator. It expects `/` literally. – Sotirios Delimanolis Apr 22 '14 at 19:36
Aha, in that case I'll change my answer then. – Rogue Apr 22 '14 at 19:37
However, I didn't down vote, so an explanation would still be nice. – Sotirios Delimanolis Apr 22 '14 at 19:39
1

The main problem is that the resource exists but sometimes cannot be found. This would treat it safely but doesn't solve the problem. – Luiggi Mendoza Apr 22 '14 at 19:40
Where is the file in context of your project? – Rogue Apr 22 '14 at 19:41

Java input == null why?

2 Answers2