On Jan 24, 2008 3:47 AM, Kamal Fariz <[EMAIL PROTECTED]> wrote: > > With 1.1.1 you can do this: > > > > Given "$n_users in the system" do |num_users| > > case num_users > > when "a user" > > # one user case > > when /(\d*) users/ > > # multi user case using $1 > > end > > end > > > > With 1.1.2 you can do this: > > > > Given /(a|\d*) users? in the system/ do |num_users| > > num_users = (num_users == 'a') ? 1 : num_users.to_i > > (1..num_users).each do > > # ... > > end > > end > > Didn't know you could do that! Thanks. So, the step will yield all the > matches in the same order regexp will populate the matches array?
Yep. Cheers, David > > > Regards, > kamal > > _______________________________________________ > rspec-users mailing list > rspec-users@rubyforge.org > http://rubyforge.org/mailman/listinfo/rspec-users > _______________________________________________ rspec-users mailing list rspec-users@rubyforge.org http://rubyforge.org/mailman/listinfo/rspec-users
