regular expression in sed for masking credit card

Question

We need to mask credit card numbers.Masking all but last 4 digits. I am trying to use SED. As credit card number length varies from 12 digits to 19,I am trying to write regular expression.Following code will receive the String. If it contains String of the form "CARD_NUMBER=3737291039299199", it will mask first 12 digits. Problem is how to write regular expression for credit card-12 to 19 digits long? If I write another expression for 12 digits, it doesn't work.that means for 12 digit credit card- first 8 digits should be masked. for 15 digit credit card, first 11 digits should be masked.

 while read data; do
    var1=${#data}

    echo "Length is "$var1
    echo $data | sed -e "s/CARD_NUMBER=\[[[:digit:]]\{12}/CARD_NUMBER=\[\*\*\*\*\*\*\*\*/g"
    done

Answer 1

How about

sed -e :a -e "s/[0-9]\([0-9]\{4\}\)/\*\1/;ta"

(This works in my shell, but you may have to add or remove a backslash or two.) The idea is to replace a digit followed by four digits with a star followed by the four digits, and repeat this until it no longer triggers.

Answer 2

This does it in one sed command without an embedded newline:

sed -r 'h;s/.*([0-9]{4})/\1/;x;s/CARD_NUMBER=([0-9]*)([0-9]{4})/\1/;s/./*/g;G;s/\n//'

If your sed doesn't have -r:

sed 'h;s/.*\([0-9]\{4\}\)/\1/;x;s/CARD_NUMBER=\([0-9]*\)\([0-9]\{4\}\)/\1/;s/./*/g;G;s/\n//'

If your sed needs -e:

sed -e 'h' -e 's/.*\([0-9]\{4\}\)/\1/' -e 'x' -e 's/CARD_NUMBER=\([0-9]*\)\([0-9]\{4\}\)/\1/' -e 's/./*/g' -e 'G' -e 's/\n//'

Here's what it's doing:

• duplicate the number so it's in pattern space and hold space
• grab the last four digits
• swap them into hold space and the whole number into pattern space
• grap all but the last four digits
• replace each digit with a mask character
• append the last four digits from hold space to the end of the masked digits in pattern space (a newline comes along for free)
• get rid of the newline