BASH：如何在watch命令中进行输出

Question

My linux 'watch' command is quite old and doesn't support '--color' option. How can I have same output like it does? because in my script the loop gives output one after another(of course). But i need it to replace the previous. Is there any tricks with terminal output?

#!/bin/bash

while true
do
/usr/sbin/asterisk -rx "show queue My_Compain" \
| grep Agent \
| grep -v \(Unavailable\) \
| sort -t"(" -k 2 \
| GREP_COLOR='01;31' egrep -i --color=always '^.*[0-9] \(Not in use.*$|$' \
| GREP_COLOR='01;36' egrep -i --color=always '^.*\(Busy*$|$'
sleep 2
done

Answer 1

You can use clear to clear the screen before dumping your output to give the appearance of in-place updates.

To reduce blinking, you can use the age old technique of double buffering:

#!/bin/bash

while true
do
  buffer=$(
    clear
    /usr/sbin/asterisk -rx "show queue My_Compain" \
    | grep Agent \
    | grep -v \(Unavailable\) \
    | sort -t"(" -k 2 \
    | GREP_COLOR='01;31' egrep -i --color=always '^.*[0-9] \(Not in use.*$|$' \
    | GREP_COLOR='01;36' egrep -i --color=always '^.*\(Busy*$|$'
  )
  echo "$buffer"
  sleep 2
done