[SOLVED] Watch not interpreting escape codes in bash script

Arashi · 07-19-2010, 10:52 AM

Hi there,

I'm fairly new to bash scripting and already having trouble. I need a script that can print some series of strings in colors based in the information of a file, for simplicity let's say it only does:

Code:

#!/bin/bash
printf "\e[1;31;32m%-10s\e[00m" "OK"

When you execute this in the command line it prints a bold green 'OK'. So far so good.

Now, I need to check the output of the script over time using the command watch. The problem then arises. watch seems to ignore the escape codes and just prints:

Code:

[1;31;32mOK [00m

Is there any way to fix this?

If not, how can I check inside the script if it is being executed from a command? (watch in this case) So I can print without color for those cases.

Thanks for your time in advance!

GrapefruiTgirl · 07-19-2010, 11:52 AM

Seems this `watch` thing is really finicky, and all I've found on the net about getting it to print colors, involves `cat -v` but I couldn't get that to print the colors either.

Here's a hacky way of not printing colors when `watch` is running, but printing colors when running the script alone. Note that this does not determine if `watch` is watching the very script we're executing here, it ONLY checks to see if `watch` is running at all. So, if you happen to be running `watch` somewhere else on the machine for some other task, then THIS program will still not print in color.

I could not figure out a way to determine if BOTH `watch` was running AND was watching this very program. There's probably a way, but I haven't found it yet.

Code:

#!/bin/bash

if [ "$(ps -C watch --no-headers)" ] ; then
  printf "watch is running, so not in color"
else
  printf "\e[1;31;32m%-10s\e[00m" "no watch running, so using color"
fi

smoker · 07-19-2010, 01:09 PM

Try putting the OK inside the first set of quotes.

Code:

printf "\e[1;31;32m%-10s\e[00mOK"

or

echo -e "\E[0;32mOK"; tput sgr0

It doesn't fix the watch problem though.

Quote:

man watch:

BUGS
...
Non-printing characters are stripped from program output. Use
"cat -v" as part of the command pipeline if you want to see
them.

Unfortunately, cat -v displays ALL the non-printing characters and still no colour.

Code:

[smoker@kids ~]$ watch -n 10 ./tput.sh | cat -v
^[[?1049h^[[1;24r^[[m^[(B^[[4l^[[?7h^[[H^[[2JEvery 10.0s: ./tput.sh^[[1;57HMon Jul 19 19:46:14 2010^[[3;1H[1;31;32mOK^[[3;20H[00m^M^[[4d[0;32mOK^M^[[5d[m(B^[[24;80H[smoker@kids ~]$

Kenhelm · 07-19-2010, 01:50 PM

It's possible to bypass the watch command by sending the script output directly to the terminal.
This works on my system

Code:

#!/bin/bash
printf "\r\e[1;31;32m%-10s\e[00m" "OK $(date)" > /dev/tty

The carriage return '\r' is to keep the output at the start of the line.
The $(date) is added just to show it's working every 2 seconds.
This method can sometimes interfere with the watch command header line; the header can be suppressed to give neater output by using the '-t' option:
watch -t script.sh

Arashi · 07-27-2010, 08:31 AM

Ups... forgot to update this post!

Everyone, thanks for your answers, they helped me solving the problem.

This is how I ended up doing it.

Code:

n=`ps aux | grep 'watch' | grep 'script.sh'`	
if [ "$n" ]; then
	printf "Printing without colors..."
fi

Whit this the script can check if there is a watch calling it and, in that case, print with no colors.