UNIX for Dummies Questions & Answers

Hi.

Here is a short perl script, but note that it will fail on numbers that contain decimal points:

Code:

#!/bin/bash -

# @(#) s1       Demonstrate commify for signed integer strings in perl.

echo
echo "(Versions displayed with local utility \"version\")"
version >/dev/null 2>&1 && version =o $(_eat $0 $1) perl
echo

rm -f t1

cat <<'EOF' |
999
1000
+1000
-1000
999.999
1000.1234
We saw $25000 in the cash box.
EOF
perl -wpe '1 while s/(\d+)(\d\d\d)/$1,$2/;'

exit 0

Producing:

Code:

% ./s1

(Versions displayed with local utility "version")
Linux 2.6.11-x1
GNU bash 2.05b.0
perl 5.8.4

999
1,000
+1,000
-1,000
999.999
1,000.1,234
We saw $25,000 in the cash box.

This is similar to the code found in the Perl Cookbook, and at several places on the net. It has the advantage of being to able to do the work without extracting the integer string -- if something doesn't look like an integer, it skips over it. It will not skip leading zeros, nor add the $ for you -- just basic comma insertion.

There are pure shell functions that can do the same thing. Balancing the sizes of the codes and the size of the interpreters, I'd usually chose the sed solution. However, if one is doing something in awk or perl, one can use these methods directly in the code. I like the brevity of the perl script, made possible by regular expressions.

Additional perl solutions can be found the recent thread converting number representation - LinuxQuestions.org ... cheers, drl