I took this code straight out of 'Java all in one for Dummies' ....why doesn't it work?

Question

import javax.swing.ImageIcon;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JPanel;


@SuppressWarnings("serial")
public class Picture extends JFrame{

    public static void main(String[] args){

        new Picture();

    }

    public Picture(){

        this.setTitle("Picture");
        this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        JPanel p = new JPanel();
        ImageIcon pic = new ImageIcon("TestImage.jpg");
        p.add(new JLabel(pic));
        this.add(p);
        this.pack();
        this.setVisible(true);
        System.out.println(pic);
    }

}

This code as it is just displays a collapsed frame. I would expect that the pack(); method call would size the frame around my .jpg image, correct?

I can already hear the first question, and yes the image is in the same folder as my project package folder.

I am using the Eclipse IDE.

I also added the println(); statement to see what the Image reference returned and it returns "TestImage.jpg". I thought it would return an address of the image in memory.??

I have been reading the forum all morning and I can't really find anything that helps me there without copying and using someone else's more complex code. I am trying to keep this as simple as I can so that I don't confuse myself.

Thanks in advance for any help.

"I took this code straight out of 'Java all in one for Dummies' …why doesn't it work?" ROFL Best question title ever! Thank you :) — Geeks On Hugs, Aug 04 '12 at 22:28
Try doing `this.pack()` _after_ the `this.setVisible(true)` call. ;) — obataku, Aug 04 '12 at 22:30
To answer a minor question: `System.out.println(pic)` will print whatever text (String) the object's `.toString()` method returns. — dsh, Aug 04 '12 at 22:33
If this image is supplied by the user, give them a `JFileChooser` to identify it. If it is part of the application, use `getResource()`. — Andrew Thompson, Aug 04 '12 at 23:04
@JavainAlaska: Hopefully this [answer](http://stackoverflow.com/a/9866659/1057230), be able to add more to your knowledge in your endeavour :-) — nIcE cOw, Jul 23 '14 at 17:01

tenorsax · Answer 1 · 2012-08-04T22:55:57.100

6

Try:

ImageIcon image = new ImageIcon(getClass().getResource("TestImage.jpg"));

See How to Use Icons tutorial for more details, in particular Loading Images Using getResource section.

edited Aug 04 '12 at 22:55

answered Aug 04 '12 at 22:31

tenorsax

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Why are you calling `getClass()` twice? – obataku Aug 04 '12 at 22:34

Mechanical snail · Accepted Answer · 2012-08-04T22:49:54.700

the image is in the same folder as my project package folder

That's it.

As written, your program looks for the image in the current working directory, not the package hierarchy.

From the Javadoc for the constructor taking String, it reads the image from the specified filename, as desired. However, when you specify a relative path, that means to read relative to the working directory that the application is running in.

To work around this, you have 2 options:

Specify the image filename relative to the working directory your IDE runs your program in. I believe Eclipse runs applications in the project root directory, and the source package hierarchy is rooted at src. In this case it'll work if you specify src/TestImage.jpg. The disadvantage is that if you ever run your program from a different directory, you'll have to move the image file along with it. This is inconvenient for distribution/packaging, because you can't just drop the JAR file and have it run.
Use Java's resource loader to load the image file from the package hierarchy. To do this, first use
```
getClass().getResource("TestImage.jpg")
```
to get a URL for the image (relative to the package root). See that ImageIcon has a constructor that accepts a URL to read the image from. So you should read the image using
```
new ImageIcon(getClass().getResource("TestImage.jpg"))
```
The advantage of being relative to he package hierarchy is that the program can be run from any location, and the image can be bundled with your app in a single JAR file.

Aside: it's best practice to create a package in which you place both code and resources (rather than just placing them in the package root). In that case, pass "com/example/someapp/TestImage.jpg" instead.

actually, in eclipse, this is the *current working directory*. — Lukas Knuth, Aug 04 '12 at 22:31
+1 for the explanation of the problem. -1 for not discussing reasons to use `getResource()`. Nice catch! — Richard Sitze, Aug 04 '12 at 22:40
new ImageIcon(getClass().getResource("TestImage.jpg")); This Worked!! why would there not be some sort of exception when the file was not found? — Abstraction is everything., Aug 04 '12 at 23:07
@Mechanicalsnail I mean if the path was wrong the way I had it in the first place, then it wasn't finding the file. Right? so how come there is no File not found exception? — Abstraction is everything., Aug 04 '12 at 23:23
`new javax.swing.ImageIcon("nonexistent/file.jpg")` throws no exception, so the libraries seem to be swallowing `FileNotFoundException` somewhere. I don't know why they designed it that way. — Mechanical snail, Aug 04 '12 at 23:27

I took this code straight out of 'Java all in one for Dummies' ....why doesn't it work?

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