Say, I want to get all files in one folder. Since Dir.entries will
have "." and ".." included. I want to strip them out with "grep". I
tried to write in such a way as I can did similarly in Perl:
log_files = Dir.entries(log_dir).grep(!/^\.\.?/)
But, it is a syntax error to have "!/regex/" as grep's argument.
Should I use some trick to make grep accept "not-match" regex?
Hi,
At Tue, 5 Jun 2007 18:55:02 +0900,
Morgan Cheng wrote in [ruby-talk:254395]:
Say, I want to get all files in one folder. Since Dir.entries will
have "." and ".." included. I want to strip them out with "grep". I
tried to write in such a way as I can did similarly in Perl:
log_files = Dir.entries(log_dir).grep(!/^\.\.?/)
grep(/\A(?!\.\.?\z)/)
···
--
Nobu Nakada
Use #reject:
Dir.entries(".").reject {|x| /\A\.+\z/ =~ x}
robert
···
On 05.06.2007 11:54, Morgan Cheng wrote:
Say, I want to get all files in one folder. Since Dir.entries will
have "." and ".." included. I want to strip them out with "grep". I
tried to write in such a way as I can did similarly in Perl:
log_files = Dir.entries(log_dir).grep(!/^\.\.?/)
But, it is a syntax error to have "!/regex/" as grep's argument.
Should I use some trick to make grep accept "not-match" regex?