Hi, I have to parse a lot of bad links like: http://www.something.com/bad link.jpg (spaces in them) URI.parse fails parsing them and considers them as faulty links. However in my case I need it to work. Is there a workaround for this?
on 2009-04-13 14:54
on 2009-04-13 15:03
That's not necessarily a bad url, you just need to encode it properly, i believe %20 is the code for a space. Google gives me this summary; http://www.blooberry.com/indexdot/html/topics/urle...
on 2009-04-13 15:07
On Mon, Apr 13, 2009 at 12:54 PM, Marcelo B. <email@example.com> wrote: > > -- > M. > > uri = "http://www.something.com/bad link.jpg" URI.parse(uri.gsub(/ /, '+')) #<- replaces all spaces with '+' Andrew T. http://ramblingsonrails.com http://www.linkedin.com/in/andrewtimberlake "I have never let my schooling interfere with my education" - Mark Twain
on 2009-04-13 15:34
Thanks for both the answers. CGI.escape is my friend:)
on 2009-04-14 02:24
On Mon, Apr 13, 2009 at 12:34 PM, Marcelo B. <firstname.lastname@example.org> wrote: > Thanks for both the answers. > > CGI.escape is my friend:) > > -- > M. > > Just a small point - URI escaping and CGI escaping are similar but not the same. To get URI escaping, require 'uri' and use URI.escape instead of CGI.escape. Regards, Sean