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: 
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
try this, you don't have to create complicated regex

```
var1="CARD_NUMBER=3737291039299199"
IFS="="
set -- $var1
cardnumber=$2
echo $cardnumber | awk 'BEGIN{OFS=FS=""}{for(i=1;i<=NF-4 ;i++){ $i="*"} }1'

```

output

```
$ ./shell.sh
************9199

```