Shell Programming and Scripting

Hi All,
I have a directory containing a lot of files and I want to search all files which contains IP addresses pattern and want to replace those ip addresses with equal number of, say, some character, 'x'. So after replacement the file should not contain any IP address. Like
10.122.53.5 -> xx.xxx.xx.x
12.176.254.24 -> xx.xxx.xxx.xx
the replacement should not change anything other than those IP address numbers, retaining all white spaces, newlines etc. intact.

for fname in $(find "$DIR" -type f -print)
do
#for each file replace the IP numbers here in the file
done

How can I achieve this using commands like grep/sed/awk/tr or may be combining use of all these commands?

Please do care about the different shells working behavior like something may work in bash, but may not work in korn shell. I am using ksh for this.

Thanks in advance for the replies.
sanzee

If you are ok with a Perl solution:

Code:

$ cat file
abc 10.122.53
10.122.53.5
efg 25.a23.33.44
12.176.254.24

Perl command:

Code:

$ perl -i -ple 'if (/(.*)((\d{1,3}\.){3}\d{1,3})(.*)/){$x=$2;$x=~s/\d/x/g;$_=$1.$x.$3;}' file

File after the command:

Code:

$ cat file
abc 10.122.53
x.xxx.xx.x
efg 25.a23.33.44
x.xxx.xxx.xx

Guru.

Thanks guru for the reply,
sorry i can;t go with perl. However the replacement should be like -

Code:

$ cat file
abc 10.122.53
10.122.53.5
efg 25.a23.33.44
12.176.254.24

output-

Code:

$ cat file
abc 10.122.53
xx.xxx.xx.x
efg 25.a23.33.44
xx.xxx.xxx.xx

So it will be equal number of x's for equal number of digits in ip address.

How about this

Code:

 perl -nle 'if (/([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])/) {($f=$&)=~ s/[0-9]/x/g; print $f;}else{print}' Filename