my problem is that I create a folder(name is IconResources) under src , in IconResources there are many pictures. Directory is like this:
This is actually a very difficult question, as the ClassLoader
has to take account of the fact that you might have another folder called IconResources
on the classpath at runtime.
As such, there is no "official" way to list an entire folder from the classpath. The solution I give below is a hack and currently works with the implementation of URLClassLoader
. This is subject to change.
If you want to robust approach, you will need to use a classpath scanner. There are many of these in library form, my personal favourite is Reflections, but Spring has one built in and there are many other options.
Onto the hack.
Generally, if you getResourceAsStream on a URLClassLoader
then it will return a stream of the resources in that folder, one on each line. The URLClassLoader
is the default used by the JRE so this approach should work out of the box, if you're not changing that behaviour.
Assume I have a project with the following classpath structure:
/
text/
one.txt
two.txt
Then running the following code:
final ClassLoader loader = Thread.currentThread().getContextClassLoader();
try(
final InputStream is = loader.getResourceAsStream("text");
final InputStreamReader isr = new InputStreamReader(is, StandardCharsets.UTF_8);
final BufferedReader br = new BufferedReader(isr)) {
br.lines().forEach(System.out::println);
}
Will print:
one.txt
two.txt
So, in order to get a List
of the resources in that location you could use a method such as:
public static List getResources(final String path) throws IOException {
final ClassLoader loader = Thread.currentThread().getContextClassLoader();
try (
final InputStream is = loader.getResourceAsStream(path);
final InputStreamReader isr = new InputStreamReader(is, StandardCharsets.UTF_8);
final BufferedReader br = new BufferedReader(isr)) {
return br.lines()
.map(l -> path + "/" + l)
.map(r -> loader.getResource(r))
.collect(toList());
}
}
Now remember, these URL
locations are opaque. They cannot be treated as files, as they might be inside a .jar
or they might even be at an internet location. So in order to read the contents of the resource, use the appropriate methods on URL
:
final URL resource = ...
try(final InputStream is = resource.openStream()) {
//do stuff
}