Help with Regexp, please
in [Programming]
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From: Ron Rosenfeld on 1 May 2010 20:58 On Sat, 1 May 2010 20:19:07 -0400, "Rick Rothstein" <rick.newsNO.SPAM(a)NO.SPAMverizon.net> wrote: >Okay, I wasn't entirely sure what the "15,16" meant in the pattern. If the >numbers are always followed by a space (is that what the "\s\S" part of your >expression is for?), my code could be modified to this... > Whatever it is that follows the string of digits is undefined in the OP's specifications. It could be a space; it could be another digit; it could be an alpha character; it could be nothing. That is a problem with the specification. Also, whatever it is that precedes the string of digits is ALSO unspecified. It could even be another digit! If, as in the OP's example, the string of digits is ALWAYS preceded and followed by a space, then his regex should have been something like: \s\d{15,16}\s If he only wanted to capture the standalone string of digits, then \s(\d{15,16})\s would capture just the digits into Group 1. and, expanding on that, \s\d{15,16}\s+([\s\S]*) would also capture everything after the string of digits into group 2 except for the leading <space>'s before group 2. In my suggestion, the [\s\S] will match every character that is either a <space> or not a <space>. In other words, it captures everything. If all I wanted to do was return everything in the string that came after a 15 or 16 digit number, that was bounded by spaces, I would just replace the beginning of the string with nothing. It's much simpler, and probably faster. =================================== Option Explicit Function Part2(s As String) As String Dim re As Object Set re = CreateObject("vbscript.regexp") re.Pattern = "^[\s\S]+\s\d{15,16}\s+" Part2 = re.Replace(s, "") End Function ==================================== If I wanted to include the space prior to " Raffles Traders" as the OP did in his example, then perhaps I would use this pattern: re.Pattern = "^[\s\S]+\s\d{15,16}\b" Now this would return the unaltered string if there was no match, but we could easily test for that, depending on what the OP wanted to do in that instance. ----------------- if re.test(s) = true then part2 = re.replace(s,"") else part2 = "no pattern match" end if -------------------- --ron
From: Raj on 2 May 2010 10:52 Thanks for the solved problem and the learning about submatches that I was not aware of. Regards, Raj On May 2, 5:58 am, Ron Rosenfeld <ronrosenf...(a)nospam.org> wrote: > On Sat, 1 May 2010 20:19:07 -0400, "Rick Rothstein" > > <rick.newsNO.S...(a)NO.SPAMverizon.net> wrote: > >Okay, I wasn't entirely sure what the "15,16" meant in the pattern. If the > >numbers are always followed by a space (is that what the "\s\S" part of your > >expression is for?), my code could be modified to this... > > Whatever it is that follows the string of digits is undefined in the OP's > specifications. It could be a space; it could be another digit; it could be an > alpha character; it could be nothing. That is a problem with the > specification. > > Also, whatever it is that precedes the string of digits is ALSO unspecified. It > could even be another digit! > > If, as in the OP's example, the string of digits is ALWAYS preceded and > followed by a space, then his regex should have been something like: > > \s\d{15,16}\s > > If he only wanted to capture the standalone string of digits, then > > \s(\d{15,16})\s would capture just the digits into Group 1. > > and, expanding on that, > > \s\d{15,16}\s+([\s\S]*) would also > capture everything after the string of digits into group 2 except for > the leading <space>'s before group 2. > > In my suggestion, the [\s\S] will match every character that is either a > <space> or not a <space>. In other words, it captures everything. > > If all I wanted to do was return everything in the string that came after a 15 > or 16 digit number, that was bounded by spaces, I would just replace the > beginning of the string with nothing. It's much simpler, and probably faster. > =================================== > Option Explicit > Function Part2(s As String) As String > Dim re As Object > Set re = CreateObject("vbscript.regexp") > re.Pattern = "^[\s\S]+\s\d{15,16}\s+" > Part2 = re.Replace(s, "") > End Function > ==================================== > > If I wanted to include the space prior to " Raffles Traders" as the OP did in > his example, then perhaps I would use this pattern: > > re.Pattern = "^[\s\S]+\s\d{15,16}\b" > > Now this would return the unaltered string if there was no match, but we could > easily test for that, depending on what the OP wanted to do in that instance. > > ----------------- > if re.test(s) = true then > part2 = re.replace(s,"") > else > part2 = "no pattern match" > end if > -------------------- > --ron
From: Ron Rosenfeld on 2 May 2010 11:27
On Sun, 2 May 2010 07:52:40 -0700 (PDT), Raj <rspai9(a)gmail.com> wrote: >Thanks for the solved problem and the learning about submatches that I >was not aware of. > >Regards, >Raj Glad to help. Thanks for the feedback. In your initial posting, you had your regex within parentheses: (\d{15,16}) ^ ^ That captures that result into a capturing group, which, in VBA, is referenced as a submatch. That was why I assumed that you were aware of that concept. I should have been more explicit. --ron |