Shell Programming and Scripting

It starts off very much like your original attempt.

\|~DRG\|(\d+): Begins by looking for |~DRG|<some number>. The number is captured.

.*?: ... followed by a non-greedy wildcard, so it doesn't skip anything we don't want to skip ...

(?=\|~(?!DCT)) ... until a |~ is found that is not followed by DCT. This uses a negative look-ahead assertion nested within a positive look-ahead assertion, so that we don't consume this portion of the record, leaving it for the substitution's next iteration.

/$1>15 ? "" : $&/ge: The e flag indicates that the replacement text should be treated as a perl expression. We test if the number captured by the first parenthetical, $1, is greater than 15. If it is, everything that was matched is replaced with an empty string. If $1 is not greater than 15, $& replaces the text with itself.

Regards,
Alister

Actually I think I have it figured out. Please correct me if not.

-l = chomp each record (strips newlines)
-p = Automatically loop through each line and print it
-e = define code to be executed

s/\|~DRG\| = Search for the patterm "|~DRG|"
(\d+) = followed by a positive number (group it so we can refer to it as $1)
.*? = followed by any character, any number of characters stopping at the first match on the line of the second group:
(?=\|~(?!DCT)) = match where the pattern equals "|~" followed by a string that is NOT "DCT"
/$1>15?"":$& = replace that with the results of a test. If the first group (the positive nbr) is > 15, replace with an empty string. Else replace with the results of the last match (the segment itself).
/g = perform the replacements on every occurrence on the line
e = evaluate the replacement expression as if it is a Perl expression, rather than just a regular expression

Whew

---------- Post updated at 06:03 PM ---------- Previous update was at 06:02 PM ----------

You beat me to it! At least I was right in figuring it out.

Thanks a million!