No.

Code:

        sub(/^[^0-9]*/,"",A) # Delete prefix

deletes everything from the start of the string that is not a decimal digit.

Code:

        N=length($0)-length(A) # Measure length of prefix from this

then computes the length of the alphabetic part of your input as the original line length minus the line length of the input with the characters that are not decimal digits at the start of your input removed. With the input line "S2<space><space>", N is set to 1, i.e., 4 (4 input characters) - 3 (the length of "2<space><space>" after removing the leading "S").

Code:

        printf("%s%0" 8-N "d\n", substr($0,1,N), A)

and this works because the substr() extracts the 1st character from the original input and prints it using the format %s and prints "2<space><space>" using the format %07d which prints the 7 (or more) digit leading zero filled number specified by A (and the %d format specifier ignores anything in the string it evaluates starting with the 1st character that is not part of a valid numeric value).

My code looks for trailing digits to determine the numeric part of your input (allowing other digits to appear elsewhere in the prefix). When there aren't any trailing decimal digits:

Code:

	NUMBER = substr($0, match($0, /[[:digit:]]*$/))

saves the trailing decimal digits in the variable NUMBER and the call to match has the side effect of setting RSTART to the offset in $0 where the first decimal digit was found (zero if not match was found) and setting RLENGTH to the number of decimal digits found at the end of the $0 (-1 if no match is found).

Code:

	PREFIX = substr($0, 1, PRELEN = (RSTART - 1))

sets PREFIX to the prefix (your alphabetic part, but this will take use the longest string at the start of the line that does not end in a decimal digit). And it sets PRELEN to the length of that string.

Code:

	DIGITS = LEN - PRELEN

sets DIGITS to the length of the string you want (8) minus the length of PREFIX (PRELEN).

Code:

	printf "%s%0*d\n", PREFIX, DIGITS, NUMBER

and here we print the PREFIX saved above and uses the same %0xd format to print the decimal digits found at the end of your input.

And, with the input "S2<space><space>", what we find if we look closely is that the awk match() function on Mac OS X (from BSD) does not conform to the standards. When no match is found RSTART should be set to zero, but instead it is being set to the length of the input string plus one. So, the almost reasonable output I showed you in post #22 for the input with trailing spaces is not what a standards-conforming awk should do. (I just love it when I find conformance bugs in UNIX implementations when I'm trying to explain how things should work! ) So, now I need to check to see what other implementations do to determine if this is a bug in the standards or a bug in BSD/Apple awk. Are we having fun yet...

Update: Please ignore the grayed out paragraph above... I obviously hadn't had enough sleep when I wrote it. I'll post an update later today explaining correctly how Apple/BSD awk is doing exactly what it is supposed to be doing with the input string "S2<space><space>"... I apologize for any confusion this may have caused.