Get parts of a NSURL in objective-c

This isn't exactly the third level, mind you. An URL is split like that way:

• the protocol or scheme (here, http)
• the :// delimiter
• the username and the password (here there isn't any, but it could be username:password@hostname)
• the host name (here, digg.com)
• the port (that would be :80 after the domain name for instance)
• the path (here, /news/business/24hr)
• the parameter string (anything that follows a semicolon)
• the query string (that would be if you had GET parameters like ?foo=bar&baz=frob)
• the fragment (that would be if you had an anchor in the link, like #foobar).

A "fully-featured" URL would look like this:

http://foobar:nicate@example.com:8080/some/path/file.html;params-here?foo=bar#baz

NSURL has a wide range of accessors. You may check them in the documentation for the NSURLclass, section Accessing the Parts of the URL. For quick reference:

• -[NSURL scheme] = http
• -[NSURL resourceSpecifier] = (everything from // to the end of the URL)
• -[NSURL user] = foobar
• -[NSURL password] = nicate
• -[NSURL host] = example.com
• -[NSURL port] = 8080
• -[NSURL path] = /some/path/file.html
• -[NSURL pathComponents] = @["/", "some", "path", "file.html"] (note that the initial / is part of it)
• -[NSURL lastPathComponent] = file.html
• -[NSURL pathExtension] = html
• -[NSURL parameterString] = params-here
• -[NSURL query] = foo=bar
• -[NSURL fragment] = baz

What you'll want, though, is something like that:

NSURL* url = [NSURL URLWithString:@"http://digg.com/news/business/24hr"];
NSString* reducedUrl = [NSString stringWithFormat:
    @"%@://%@/%@",
    url.scheme,
    url.host,
    url.pathComponents[1]];

For your example URL, what you seem to want is the protocol, the host and the first path component. (The element at index 0 in the array returned by -[NSString pathComponents] is simply "/", so you'll want the element at index 1. The other slashes are discarded.)