print first 3 characters and / rest of the string with stars - bash

Question

I'have this input like this
John:boofoo
I want to print rest of the string with stars and keep only 3 characters of the string.
The output will be like this
John:boo***
this my command
awk -F ":" '{print $1,$2 ":***"}'
I want to use only print command if possible. Thanks

Answer 1

With GNU sed:
echo 'John:boofoo' | sed -E 's/(:...).*/\1***/'
Output:
John:boo***

Answer 2

With GNU awk for gensub():
$ awk 'BEGIN{FS=OFS=":"} {print $1, substr($2,1,3) gensub(/./,"*","g",substr($2,4))}' file
John:boo***
With any awk:
awk 'BEGIN{FS=OFS=":"} {tl=substr($2,4); gsub(/./,"*",tl); print $1, substr($2,1,3) tl}' file
John:boo***

Answer 3

Could you please try following. This will print stars(keeping only first 3 letters same as it is) how many characters are present in 2nd field after first 3 characters.
awk '
BEGIN{
FS=OFS=":"
}
{
stars=""
val=substr($2,1,3)
for(i=4;i<=length($2);i++){
stars=stars"*"
}
$2=val stars
}
1
' Input_file
Output will be as follows.
John:boo***
Explanation: Adding explanation for above code too here.
awk '
BEGIN{ ##Starting BEGIN section from here.
FS=OFS=":" ##Setting FS and OFS value as : here.
} ##Closing block of BEGIN section here.
{ ##Here starts main block of awk program.
stars="" ##Nullifying variable stars here.
val=substr($2,1,3) ##Creating variable val whose value is 1st 3 letters of 2nd field.
for(i=4;i<=length($2);i++){ ##Starting a for loop from 4(becasue we need to have from 4th character to till last in 2nd field) till length of 2nd field.
stars=stars"*" ##Keep concatenating stars variable to its own value with *.
}
$2=val stars ##Assigning value of variable val and stars to 2nd field here.
}
1 ##Mentioning 1 here to print edited/non-edited lines for Input_file here.
' Input_file ##Mentioning Input_file name here.

Answer 4

Or even with good old sed
$ echo "John:boofoo" | sed 's/...$/***/'
Output:
John:boo***
(note: this just replaces the last 3 characters of any string with "***", so if you need to key off the ':', see the GNU sed answer from Cyrus.)

Answer 5

Another awk variant:
awk -F ":" '{print $1 FS substr($2, 1, 3) "***"}' <<< 'John:boofoo'
John:boo***

Answer 6

Since we have the tags awk, bash and sed: for completeness sake here is a bash only solution:
INPUT="John:boofoo"
printf "%s:%s\n" ${INPUT%%:*} $(TMP1=${INPUT#*:};TMP2=${TMP1:3}; echo "${TMP1:0:3}${TMP2//?/*}")
It uses two arguments to printf after the format string. The first one is INPUT stripped of by everything uncluding and after the :. Lets break down the second argument $(TMP1=${INPUT#*:};TMP2=${TMP1:3}; echo "${TMP1:0:3}${TMP2//?/*}"):
$(...) the string is interpreted as a bash command its output is substituted as last argument to printf
TMP1=${INPUT#*:}; remove everything up to and including the :, store the string in TMP1.
TMP2=${TMP1:3}; geht all characters of TMP1 from offset 3 to the end and store them in TMP2.
echo "${TMP1:0:3}${TMP2//?/*}" output the temporary strings: the first three chars from TMP1 unmodified and all chars from TMP2 as *
the output of the last echo is the last argument to printf
Here is the bash -x output:
+ INPUT=John:boofoo
++ TMP1=boofoo
++ TMP2=foo
++ echo 'boo***'
+ printf '%s:%s\n' John 'boo***'
John:boo***

Answer 7

Another sed : replace all chars after the third by *
sed -E ':A;s/([^:]*:...)(.*)[^*]([*]*)/\1\2\3*/;tA'

Answer 8

Some more awk
awk 'BEGIN{FS=OFS=":"}{s=sprintf("%0*d",length(substr($2,4)),0); gsub(/0/,"*",s);print $1,substr($2,1,3) s}' infile
You can use the %* form of printf, which accepts a variable width. And, if you use '0' as your value to print, combined with the right-aligned text that's zero padded on the left..
Better Readable:
awk 'BEGIN{
FS=OFS=":"
}
{
s=sprintf("%0*d",length(substr($2,4)),0);
gsub(/0/,"*",s);
print $1,substr($2,1,3) s
}
' infile
Test Results:
$ awk --version
GNU Awk 3.1.7
Copyright (C) 1989, 1991-2009 Free Software Foundation.
$ cat f
John:boofoo
$ awk 'BEGIN{FS=OFS=":"}{s=sprintf("%0*d",length(substr($2,4)),0); gsub(/0/,"*",s);print $1,substr($2,1,3) s}' f
John:boo***

Answer 9

Another pure Bash, using the builtin regular expression predicate.
input="John:boofoo"
if [[ $input =~ ^([^:]*:...)(.*)$ ]]; then
printf '%s%s\n' "${BASH_REMATCH[1]}" "${BASH_REMATCH[2]//?/*}"
else
echo >&2 "String doesn't match pattern"
fi
We split the string in two parts: the first part being everything up to (and including) the three chars found after the first colon (stored in ${BASH_REMATCH[1]}), the second part being the remaining part of string (stored in ${BASH_REMATCH[2]}). If the string doesn't match this pattern, we just insult the user.
We then print the first part unchanged, and the second part with every character replaced with *.