ANSI escape codes for coloring inside bash printf
bash, shell
Solution
You're looking for a format specifier that will expand escape characters in the argument. Conveniently, bash supports (from `help printf`):
%b expand backslash escape sequences in the corresponding argument
Alternatively, bash also supports a special mechanism by which will perform expansion of escape characters:
d=$'\e[33m'
Problem
Lines `8.` and `9.` below confound me: ``` #!/bin/bash a=foo b=6 c=a d="\e[33m" # opening ansi color code for yellow text e="\e[0m" # ending ansi code f=$d printf "1. foo\n" printf "2. $a\n" printf "3. %s\n" "$a" printf "4. %s\n" "${!c}" printf "5. %${b}s\n" "$a" printf "6. $d%s$e\n" "$a" # will be yellow printf "7. $f%s$e\n" "$a" # will be yellow printf '8. %s%s%s\n' "$d" "$a" "$e" # :( printf "9. %s%s%s\n" "$f" "$a" "$e" # :( ``` Is it possible to use `%s` to expand a colour variable and see the colour switch? Output: ``` 1. foo 2. foo 3. foo 4. foo 5. foo 6. foo 7. foo 8. \e[33mfoo\e[0m 9. \e[33mfoo\e[0m ``` Note: `6.` and `7.` are indeed yellow Edit ``` printf "10. %b%s%b\n" "$f" "$a" "$e" # :) ``` ... finally! That's the command that does it, thanks to Josh!