UNIX for Dummies Questions & Answers

A developer of mine has this requirement - I couldn't tell her quickly how to do it with UNIX commands or a quick script so she's writing a quick program to do it - but that got my curiousity up and thought I'd ask here for advice.

In a text file, there are some records (about half of them) that have a specific string, say "ABC" followed by a 15 digit number, always at least 2 leading zeros. In rows that have this, it will appear twice, identically.
I essentially want to cut out these 18 chars into a file of their own. But, they are not in a fixed column position within the file.

Logically, the task is:
a) find the rows with ABC00
b) get the position of that first A
c) cut starting at that position for 18 characters and write to a new file.

example data:
ab cdefgABC000000000012345ABC000000000012345sadlfk
abcde fgABC000000000012346ABC000000000012346sadlfk
abc defgghi jklmn1349d5sadlfk
abcdef sldkfdgABC000000000056789ABC000000000056789abcdlkdfj134239d

and so on.

Desired output
ABC00000000012345
ABC00000000012346
ABC00000000056789

Thanks for having a look.
Lisa

Lisa,
There are probably many, but here is one approach --

Code:

> sed "s/ABC[0-9][0-9]/~+&/" file220 | tr "~" "\n" | grep "+" | cut -c2-19
ABC000000000012345
ABC000000000012346
ABC000000000056789

wow! that's slick - and it worked on my data stream so 1000s of thanks. Now, the ethical dilemma, do I just give it to the developer as if I did it or do I 'fess up that I asked for help.

Lisa

You found a solution and verified it works.

Most every problem has already been pondered and solved, so there truly are no "new" answers. Ha ha

Back to the initial problem, the creative use of sed to place extra characters and then tr to convert them so a grep and cut can extract them -- is one useful process to pull apart data records.

Let him think you were the genius.

Inevitably, a perl approach

Code:

perl -ne '/(ABC00\d{13})/ && print "$1\n"' list.txt