27 Oct 09 BASH Script – Blank Out CC Details

Edit: I should have pointed out originally, as I have now received feedback on this. This is NOT the best or optimal way of performing this task. I was trying to illustrate as many shell scripting principles as possible in terms of ‘if’, ‘for’, counters, etc, and how such a one liner has been put together. Perhaps I should have thought of a better way of illustrating such principles, but nevertheless, here it is!

Here’s a quick one liner, can’t think why anyone would ever have any use for it, but maybe the principle itself could be of use to someone! This will take a file containing listings of 16 digit numbers, i.e. 1234123412341234 and replace it with XXXXXXXXXXXX1234

Duly spaced and indented:

Would love anyone to comment with variations.

Tags: bash, echo, grep, script, tr

Reader's Comments

1. Wesley Shields | 28 Oct 2009 2:54 pm

   I’d just do it with a regular expression…

   wxs@rst wxs % cat foo
   a
   a111111111112222
   1a11111111112222
   1111111111112222
   222222222223333
   33333333333334444
   wxs@rst wxs % sed -e ’s/^[0-9]\{12\}\([0-9]\{4\}\)$/XXXXXXXXXXXX\1/’ < foo
   a
   a111111111112222
   1a11111111112222
   XXXXXXXXXXXX2222
   222222222223333
   33333333333334444
   wxs@rst wxs %

2. Carsten | 29 Oct 2009 3:01 pm

   I’m with Wesley here. Why iterate with (poorly performing) bash, when you can use faster Perl Regular Expressions in a single command instead?

   Using perl instead of sed, you can get the regex even shorter (and it’s platform-independent as sed takes different parameters on Linux and OSX for extended regular expressions).

   Example here: http://pastebin.com/f12cc8612

3. Jadu Saikia | 07 Dec 2009 5:48 pm

   How about this ?

   $ echo “123456789″ | sed -r “s/^(.{0})(.{5})/\1XXXXX/”

   Output:
   XXXXX6789

   http://unstableme.blogspot.com/2008/01/substitute-character-by-position-sed.html

   // Jadu

Leave a Comment