实例化java.io.File以从类路径中读取资源

Question

Supposing that I've a project structure as follows:

+ src
---+ main
------+ java
------+ resources

How can I define a java.io.File instance that is able to read an xml from resources folder?

Answer 1

It depends on what your program's working directory is.

You can get your working directory at runtime via System.getProperty("user.dir");

Here's a link with more info

Once you've got that, create a relative filepath pointing to your resource folder.

For example, if your working directory is src, create a relative filepath like so:

new File(Paths.get("main","resources").toString());

There are weaknesses with this approach as the actual filepath you use may change when you deploy your application, but it should get you through some simple development.

Answer 2

No, use an InputStream, with the path starting under resources.

InputStream in = getClass().getResourceAsStrem("/.../...");

Or an URL

URL url = getClass().getResource("/.../...");

This way, if the application is packed in a jar (case sensitive names!) it works too. The URL in that case is:

"jar:file://... .jar!/.../..."

If you need the resource as file, for instance to write to, you will need to copy the resource as initial template to some directory outside the application, like System.getProperty("user.home") .

Path temp = Files.createTempFile("pref", "suffix.dat");
Files.copy(getClass().getResourceAsStream(...), temp, StandardCopyOption.REPLACE_EXISTING);
File file = temp.toFile();

As @mjaggard commented, createTempFile creates an empty file, so Files.copy will normally fail unless with option REPLACE_EXISTING.