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?
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Answer
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.