re = /^0[12][VM](?:[^\n]|\n(?!0[12][VM]|50|90))+/mi
str = '00 date
01Mxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
01 01 01 01=0xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
01=5xxxxxxxxxxxxxxxxxxxxxxxxxxx
01Mxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
01 01 01 01=0xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
01=9xxxxxxxxxxxxxxxxxxxxxxxxxxx
50 xxxxxxxxxxxxx xxxxxxxxxxxxxxxxx
01Vxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
01$1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
01$A xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
01$B 0xxxxxxxxxxxxxxxxxxxx
01$0xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
01$5xxxxxxxxxxxxxxxxxxxxxxxxxxx
50 xxxxxxxxxxxx BatchTotal
90 xxxxxxxxxxxx FILETotal'
# 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