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@Kingcobra: could you elaborate what a consecutive space means to you. Is your input file is incorrect? What have you tried?


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@Ravindersingh that is not a solution...

Hello Kingcobra,

Following may help you in same.


echo "Wow what a wonderful feeling to be alive" | awk '{for(i=1;i<=NF;i++){if(i==4){print "HRUHRUHRUHRUHRUHRUHRUWRU"} {if(i==5){print...

Try:
sed -e 's/\( *\) /\1WRU /g' -e :a -e 's/ \([HW]RU\)/ HRU\1/;ta' file

I believe there should be six HRU's (2nd until last -1) not 7, correct?

use printf "%c" "$char"
printf "$char" will break if there is "%" character read.

while IFS="" read -r -n1 chardoes the trick.
BTW echo -e and echo -n are not portable, better get used to printf:
printf "%c" "$char"
or
printf "$char"

The attached updated version should work. It does use multi-byte characters and some awk implementations may not like this, I've only tested with GNU awk.

I've replaced to attached file with a new version that avoids using UTF8 characters in awk so it should be more compatible with older awk implementations. However it looks like GNU sed is also...

Well using ls was just an example, the reason --color is needed for the ls command is that it normally detects that the output isn't a terminal and avoids outputting any color escape sequences. This...

If I am reading your post correctly then you are looking for the attributes to a string/character at any random part of an echo to a terminal.
This can't be done so you have no choice but to do it...