Extract file name from Java URL (file: and http/https protocol)?-CodePudding

I have various urls like this:

String a = "file:./bla/file.txt"; // Valid, see See [RFC 3986][1], path - rootless definition
String b = "file:.file.txt";      // Valid, see See [RFC 3986][1], path - rootless definition
String c = "file:./file.txt";     // Valid, see See [RFC 3986][1], path - rootless definition
String d = "file:///file.txt";
String e = "file:///folder/file.txt";
String f = "http://example.com/file.txt";
String g = "https://example.com/file.txt";

These are all valid URLS, and I can convert them to a URL in java without errors:

URL url = new URL(...);

I want to extract the filename from each of the examples above, so I'm left with just:

file.txt

I have tried the following, but this doesn't work for example b above (which is a valid URL):

b.substring(path.lastIndexOf('/')   1); // Returns file:.file.txt

I can prob write some custom code to check for slashes, just wondering if there a better more robust way to do it?

CodePudding user response：

The URI class properly parses the parts of a URI. For most URLs, you want the path of the URI. In the case of a URI with no slashes, there won’t be any parsing of the parts, so you’ll have to rely on the entire scheme-specific part:

URI uri = new URI(b);
String path = uri.getPath();
if (path == null) {
    path = uri.getSchemeSpecificPart();
}
String filename = path.substring(path.lastIndexOf('/')   1);

The above should work for all of your URLs.