What steps will reproduce the problem?
When calling the showUser() function and only including the id, the query
string is malformed.
I think the following would work out better:
$request = "http://twitter.com/users/show/{$qs}.{$this->type}";
Original issue reported on code.google.com by bsze...@gmail.com on 18 Mar 2009 at 3:31
Original issue reported on code.google.com by
bsze...@gmail.com
on 18 Mar 2009 at 3:31