Question

String arg="http://www.example.com/user.php?id=<URLRequest Method='GetByUID' />";
java.net.URI uri = new java.net.URI( arg );
java.awt.Desktop desktop = java.awt.Desktop.getDesktop();
desktop.browse( uri );

I want to open the given link in default browser with the above code but it says the url is invalid…i tried escaping characters like ‘ also but its not working.
If i replace String arg=”www.google.com”; then there is no problem and I am able to open google.com.
Please help.

Answer 1

Your string contains characters that aren’t valid in a URI, per RFC 2396. You need to properly encode the query parameters. Many utilities support that, like the standard URLEncoder (lower level), JAX-RS UriBuilder, Spring UriUtils, Apache HttpClient URLEncodedUtils and so on.

Edit: Oh, and the URI class can handle it, too, but you have to use a different constructor:

URI uri = new URI("http", "foo.com", null, "a=<some garbage>&b= |{$0m3 m0r3 garbage}| &c=imokay", null);
System.out.println(uri);

Outputs:

http://foo.com?a=%3Csome%20garbage%3E&b=%20%7C%7B$0m3%20m0r3%20garbage%7D%7C%20&c=imokay

which, while ugly, is the correct representation.