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    <p>You are specifying resource files with paths. That only works if you actually know the full path or the relative path from current directory; e.g. prepended with "src" as suggested in another answer.</p> <p>It is, however, much better to include resources in your build (whether you build through your IDE or an external build system like gradle or maven) so that these files are copied into the classpath (it may be that you're already doing this). In this case, you can retrieve the resources using the Java classloader, like this:</p> <pre><code>URL resource = getClass().getResource("/path/to/resource"); </code></pre> <p>A couple of notes here: </p> <ol> <li>The leading "/" specifies that the classloader should search from the root package. If it is ommitted, it will search relative from the package of the class you're in.</li> <li>You're getting a URL instead of a File. Quite often, it is possible to use these URL's directly, like in the <code>ImageIcon</code> contructor that you're using. One additional benefit with this method is that it works if your application is packaged in jar file as well, since then the URL will point into the correct location in the jar file. However, if you <em>know</em> it is a file, <em>and</em> you need the file reference, it can be converted like this: <code>File file = new File(resource.toURI());</code>. </li> </ol> <p>EDIT: The following is an example that should work for Danny, based on his "answer" post with an example.</p> <pre><code>import java.io.BufferedReader; import java.io.File; import java.io.FileReader; import java.io.IOException; import java.net.URISyntaxException; import java.net.URL; public class ResourceTest { public static void main(String[] args) throws IOException, URISyntaxException { new ResourceTest().test(); } public void test() throws URISyntaxException, IOException { URL url = getClass().getClassLoader().getResource("files/1.txt"); System.out.println("URL: " + url); File file = new File(url.toURI()); System.out.println("FILE: " + file); File folder = file.getParentFile(); System.out.println("FOLDER: " + folder); System.out.println("FOLDER LIST: "); for (String fileName : folder.list()) { System.out.println(" " + fileName); } System.out.println("FILE CONTENTS:"); BufferedReader reader = new BufferedReader(new FileReader(file)); String line = reader.readLine(); while (line != null) { System.out.println(line); line = reader.readLine(); } } } </code></pre> <p>Which for yields the following output:</p> <pre><code>URL: file:/home/steinar/Safe/source/private/out/production/files/1.txt FILE: /home/steinar/Safe/source/private/out/production/files/1.txt FOLDER: /home/steinar/Safe/source/private/out/production/files FOLDER LIST: 1.txt FILE CONTENTS: This is my test text file </code></pre>
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    1. PONot able to find files and images under src directory
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    1. COhere is what I tried but I did not work: URL url = null; Random rand = new Random(); int random = rand.nextInt(19); String filePath = "files/" + random + ".txt"; System.out.println("file: " + filePath); url = this.getClass().getClassLoader().getResource(filePath); System.out.println(url.toString().trim()); File file = new File("."); for(String fileNames : file.list()) { System.out.println(fileNames); } help please
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    2. COsorry for the messy format. I could not press the Enter key to make new line. If I press the Enter key, It just automatically save the comment and did not make new line.
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      1. USDanny Phung
    3. COWell, you're saying it didn't work, but you're not saying what happened. Your code example contains a mixture of choosing random files, resolving a URL and also listing a file path. Can you tell the the output of this: `System.out.println(getClass().getClassLoader().getResource("files/1.txt"));` If this is `null` then the resource is not copied into the classpath; then you'll need to tell me how you are running (command line? in your IDE? which tool / commands do you use). I'm assuming here that the file 1.txt exists and is in the files folder in your source directory. Is that the case?
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