re = /^(?:\d{1,10}|(?![\d.]{11,})\d+\.\d+)$/m
str = '1
12
12345
1234567890
12345678901
1.23
3.14159
1.12345678
1.123456789'
# Print the match result
str.scan(re) do |match|
puts match.to_s
end
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Ruby, please visit: http://ruby-doc.org/core-2.2.0/Regexp.html