How can I save a file to the class path

Question

How can I save / load a file that is located where my classes are? I don't the physical path to that location before and I want dynamically to find that file.

Thanks

Edit:

I want to load an XML file and write and read to it and i am not sure how to address it.

Answer 1

In the general case you cannot. Resources loaded from a classloader can be anything: files in directories, files embedded in jar files or even downloaded over the network.

Answer 2

Use ClassLoader#getResource() or getResourceAsStream() to obtain them as URL or InputStream from the classpath.

ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
InputStream input = classLoader.getResourceAsStream("com/example/file.ext");
// ...

Or if it is in the same package as the current class, you can also obtain it as follows:

InputStream input = getClass().getResourceAsStream("file.ext");
// ...

Saving is a story apart. This won't work if the file is located in a JAR file. If you can ensure that the file is expanded and is writable, then convert the URL from getResource() to File.

URL url = classLoader.getResource("com/example/file.ext");
File file = new File(url.toURI().getPath());
// ...

You can then construct a FileOutputStream with it.

Related questions:

• getResourceAsStream() versus FileInputStream

Answer 3

You can try the following provided your class is loaded from a filesystem.

String basePathOfClass = getClass()
   .getProtectionDomain().getCodeSource().getLocation().getFile();

To get a file in that path you can use

File file = new File(basePathOfClass, "filename.ext");

Answer 4

new File(".").getAbsolutePath() + "relative/path/to/your/files";

Answer 5

This is an expansion on Peter's response:

If you want the file in the same classpath as the current class (Example: project/classes):

URI uri = this.getClass().getProtectionDomain().getCodeSource().getLocation().toURI();
File file = new File(new File(uri), PROPERTIES_FILE);
FileOutputStream out = new FileOutputStream(createPropertiesFile(PROPERTIES_FILE));
prop.store(out, null);

If you want the file in a different classpath (Example: progect/test-classes), just replace this.getClass() with something like TestClass.class.

Read Properties from Classpath:

Properties prop = new Properties();

System.out.println("Resource: " + getClass().getClassLoader().getResource(PROPERTIES_FILE));
InputStream in = getClass().getClassLoader().getResourceAsStream(PROPERTIES_FILE);
if (in != null) {
    try {
        prop.load(in);
    } finally {
        in.close();
    }
}

Write Properties to Classpath:

Properties prop = new Properties();
prop.setProperty("Prop1", "a");
prop.setProperty("Prop2", "3");
prop.setProperty("Prop3", String.valueOf(false));

FileOutputStream out = null;
try {
    System.out.println("Resource: " + createPropertiesFile(PROPERTIES_FILE));
    out = new FileOutputStream(createPropertiesFile(PROPERTIES_FILE));
    prop.store(out, null);
} finally {
    if (out != null) out.close();
}

Create the File Object on the Classpath:

private File createPropertiesFile(String relativeFilePath) throws URISyntaxException {
    return new File(new File(this.getClass().getProtectionDomain().getCodeSource().getLocation().toURI()), relativeFilePath);
}

Answer 6

According to system properties documentation, you can access this as the "java.class.path" property:

string classPath = System.getProperty("java.class.path");