Scramble solution

varun_751980 · #1 (**permalink**) 12-08-08, 09:31

I have a file in UNIX. Each record in the file is 852 characters.

As per my requirement I have to replace the contents of file in each record from position 38 to 348 as “*”. These fields will have very sensitive data and we want to hide it with *.

I tried sed using –r option but the –r option doesn’t work.

Could you people please suggest something using AWK , SED or whatever. This is a urgent requirement.

stolze · #2 (**permalink**) 12-08-08, 12:47

Untested:

Code:

sed -e 's/\(.........38x....\).*/\1***********310x*****/g'

Or you can cut the first 38 characters from each line and concatenate it with 310x *.

varun_751980 · #3 (**permalink**) 12-08-08, 12:57

cThanks for the reply. But I don't want to use 310 asterik * signs... Can some command or loop help me in this

pdreyer · #4 (**permalink**) 12-09-08, 05:55

This what you want to do?
Lets assume you need to replace columns 15-20 with a single *

Code:

cat xx
ABCDEFGHIJKLMNOPQRSTUVWXYZ
----+----1----+----2----+-

cut -c 1-15,21- <xx | sed 's/./*/15'
ABCDEFGHIJKLMN*UVWXYZ
----+----1----*----+-

LKBrwn_DBA · #5 (**permalink**) 12-09-08, 11:49

Or maybe this?:

Code:

$ beg=15
$ end=20
$ echo "ABCDEFGHIJKLMNOPQRSTUVWXYZ"|\
> awk -v r="$beg,$end" 'BEGIN {split(r,q,",");for (i=q[1];i<=q[2];++i) a=a"*";}
> {o=substr($0,1,q[1]-1) a substr($0,q[2]+1); print o;}' 
ABCDEFGHIJKLMN******UVWXYZ

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