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.
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.
Thats because it is invalid. <URLRequest Method='GetByUID' /> should be replaced by the value of the id, or an expression that returns the id which you can concatenate with the arg string. Something like
String arg="http://www.example.com/user.php?id="+getByUID(someUid);
import java.net.URL;
import java.net.URLEncoder;
class ARealURL {
public static void main(String[] args) throws Exception {
String s1 = "http://www.example.com/user.php?id=";
String param = "<URLRequest Method='GetByUID' />";
String encodedParam = URLEncoder.encode(param,"UTF-8");
URL url = new URL(s1+encodedParam);
System.out.println(url);
}
}
http://www.example.com/user.php?id=%3CURLRequest+Method%3D%27GetByUID%27+%2F%3E
+. What if I want spaces encoded as %20 ?