Why isn't bash accepting my color coding?

Question

I've got a simple code setup, but for whatever reason it refuses to operate how it should

#!/bin/bash
RED='\033[0:31m'
RESET='\033[0m'

Basically these are some simple color encodings

Next thing I have is

for file in ./dir/*.c; do
echo "File [${RED}$file${RESET}] has been launched"

My expected result would be

File [myprogram.c] has been launched (while the name would be in red color)

Instead it just refuses to encode the color and dumps

File [\033[0:31m] has been launched

Any idea what I'm doing wrong?

use echo -e instead of echo, also use \033[0;31m, not colon. — ymonad
– ymonad, Commented Dec 5, 2016 at 1:16
Don't use echo to print escape sequences. As there is no portable way to do it. Use printf instead. — Andreas Louv
– Andreas Louv, Commented Dec 5, 2016 at 1:16

score 4 · Accepted Answer · 2016-12-05 02:27:09Z

4

First because red is '\033[31m'. What you wrote means something else, read below.
Then, because the codes need to be interpreted.
This won't work:

$ red='\033[31m'

$ echo "${red}Hello"

But this would:

$ red='\033[31m'

$ echo -e "${red}Hello"

Or, you can assign the interpreted values to a variable:

$ red="$(echo -e '\033[31m')"
$ red="$(printf '\033[31m')"

Then the escapes don't need to be interpreted:

$ echo "${red}Hello"

What you wrote (besides using a colon where a semicolon should be) was setting the "boldness" of the foreground, or 0 for thin/light and 1 for bold/bright.

$ printf '\033[31mHello\033[0;31mHello\033[1;31mHello'

Also, 3x is for foreground 4x is for background:

$ printf '\033[31;42mHello\033[0;44;31mHello\033[1;43;31mHello\033[0m'

A full table could be printed with this:

$ printf "$(printf '%s' 033[{0,1}';'3{1..8}{';'4{1..8}mXXX,';'40m=OoO\\033[0m\\n}  )"

edited Dec 5, 2016 at 2:27

answered Dec 5, 2016 at 1:22

user8017719

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2 Comments

Andreas Louv Over a year ago

You should really use tput to get color codes for the shell.

user8017719 Over a year ago

@andlrc Why? Doesn't the code I wrote work in your system?

V. Michel · Accepted Answer · 2016-12-05 01:28:55Z

1

With tput

#!/bin/bash
red=$(tput setaf 1)
reset=$(tput sgr0)
for file in ./dir/*.c; do
 echo "File [${red}$file${reset}] has been launched"
done

with printf

#!/bin/bash
RED='\033[0;31m'
RESET='\033[0m'
for file in ./dir/*.c; do
 printf  "File [${RED}$file${RESET}] has been launched"
done

edited Dec 5, 2016 at 1:28

answered Dec 5, 2016 at 1:22

V. Michel

1,62713 silver badges14 bronze badges

Comments

Community · Accepted Answer · 2017-04-13 12:36:24Z

There is a typo in your code. Red color code is [031m. The ASCII codes always start with an escape character, octal 33: \033. Thus, to start printing red text join the two sequences: \033[031m.

I find this post very helpful. In particular, the author recommends to use tput instead of the hard-coded values, and I agree with him.

Example

die() {
    local message="$1"
    : ${message:=Aborted}

    # See info bash BASH_SOURCE, info bash FUNCNAME, info bash BASH_LINENO
    printf '%s at %s:%s line %d\n' \
      "$message" ${BASH_SOURCE[1]} ${FUNCNAME[1]} ${BASH_LINENO[0]} >&2

    exit 1
}

# See man 1 tput, man 5 terminfo.
red=$(tput setaf 1) || die
noattr=$(tput sgr0) || die

# If tput is unavailable (very unlikely), use the hardcoded values as follows.
# The ANSI codes always start with the escape character (octal 33).
#esc='\033'
#red="${esc}[031m"
#noattr="${esc}[0m"

# %b causes printf to expand backslash escape sequences. See info bash printf.
printf '%b%s%b\n%s\n' "$red" 'red text' "$noattr" 'normal text'

Note the use of printf. Don't use echo in new software.

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