Is it possible to write a regular expression that matches a particular pattern and then does a replace with a part of the pattern

Question

I\'m working with some comma delimited text files. The file is comprised of approximately 400 rows and 94 columns all comma delimited and withing double quotes:

Answer 1

Use a look around

(?<!"),(?!")

replacing it with a pipe.

which means

(?<!")    - character before is not a "
,         - match a comma
(?!")     - character after is not a "

Answer 2

Rather than interfere with what is evidently source data, i.e. the stuff inside the quotes, you might consider replacing the field-separator commas instead:

s/,([^,"]*|"[^"]*")(?=(,|$))/|$1/g

Note that this also handles non-quoted fields.

On this data: "H",9,"YES","NO","4,5","Y","N"

$ perl -pe 's/,([^,"]*|"[^"]*")(?=(,|$))/|$1/g' commasep
"H"|9|"YES"|"NO"|"4,5"|"Y"|"N"

Which can afterwards be split on "|":

$ perl -ne 's/,([^,"]*|"[^"]*")(?=(,|$))/|$1/g;print join "---",split "\\|"' commasep
"H"---9---"YES"---"NO"---"4,5"---"Y"---"N"