[SOLVED] How to add a space after special characters in bash script?

Issue

I have a text file with something like,

!aa
@bb
#cc
$dd
%ee

expected output is,

! aa
@ bb
# cc
$ dd
% ee

What I have tried, echo "${foo//@/@ }".

This does work fine with one string but it does not work for all the lines in the file. I have tried with this while loop to read all the lines of the file and do the same using echo but it does not work.

while IFS= read -r line; do
    foo=$line
    sep="!@#$%"
    echo "${foo//$sep/$sep }"
done < $1

I have tried with awk split but it does not give the expected output. Is there any workaround for this? by using awk or sed.

Solution

The following assumes you want to add a space after every character in the !@#$% set (even if it is the last character in a line). Test file:

$ cat file.txt
a!a
@bb
c#c
$dd
ee%
foo
%b%r
$ sep='!@#$%'

With sed:

$ sed 's/['"$sep"']/& /g' file.txt
a! a
@ bb
c# c
$ dd
ee% 
foo
% b% r

With awk:

$ awk '{gsub(/['"$sep"']/,"& "); print}' file.txt
a! a
@ bb
c# c
$ dd
ee% 
foo
% b% r

With plain bash (not recommended, it is too slow):

$ while IFS= read -r line; do
    str=""
    for (( i=0; i<${#line}; i++ )); do
      char="${line:i:1}"      
      str="$str$char"
      [[ "$char" =~ [$sep] ]] && str="$str "
    done
    printf '%s\n' "$str"
  done < file.txt
a! a
@ bb
c# c
$ dd
ee% 
foo
% b% r

Or (not sure which is the worst):

$ while IFS= read -r line; do
    for (( i=0; i<${#sep}; i++ )); do
      char="${sep:i:1}"
      line="${line//$char/$char }"
    done
    printf '%s\n' "$line"
  done < file.txt
a! a
@ bb
c# c
$ dd
ee% 
foo
% b% r

Answered By - Renaud Pacalet

Answer Checked By - Gilberto Lyons (WPSolving Admin)