Regular Expressions: parsing through multiple captures

Question

Good morning. I have the following line returned from a device

♥☺i20100☺i2010018032809120110000074653B0F4465275D245EAA80042D82F5E000000004291D83C00000000022000007460042FD45FE7F2A4648C0004282FC0400000000429799E500000000033000007466A869946694E8145BCF80042EF811B00000000428CCAF400000000044000007461DFFC4461D1D61462B040042A15A6600000000428E635100000000&&C631♥

The data comes back in blocks of 6 characters after the date-time stamp. I'm trying to get past the date-time and look for all blocks of 6 that begin with 42. There should be 3 blocks returned.
Using Expresso, I can get all the blocks of .*i20100.i20100(?<Date>.{10})(?<AANNTT>.{6})+

but how can I make a subquery on each of those and return all the ones starting with 42? I thought it might be this: .*i20100.i20100(?<Date>.{10})(.{6})*(?<AANNTT>42.{4})+

but that's only returning the last one.

Thanks in advance! -Dave

Answer 1

You may use

(?<=i20100.i20100(?<Date>.{10})(?:.{6})*?)42.{4}

See the regex demo

Details

• (?<=i20100.i20100(?<Date>.{10})(?:.{6})*?) - a positive lookahead that matches a location in a string that is immediately preceded with
  • i20100.i20100 - i20100, and char but a newline, i20100
  • (?<Date>.{10})(?:.{6})*? - Group Date that matches 10 chars other than a newline, then any 0+ repetitions of any 6 chars but a newline, as few as possible
• 42.{4} - 42 and then any 4 chars other than a newline