How to get the cursor position in bash?

Question

In a bash script, I want to get the cursor column in a variable. It looks like using the ANSI escape code {ESC}[6n is the only way to get it, for example the following way:

# Query the cursor position echo -en '\033[6n'  # Read it to a variable read -d R CURCOL  # Extract the column from the variable CURCOL="${CURCOL##*;}"  # We have the column in the variable echo $CURCOL 

Unfortunately, this prints characters to the standard output and I want to do it silently. Besides, this is not very portable...

Is there a pure-bash way to achieve this ?

Answer 1

You have to resort to dirty tricks:

#!/bin/bash # based on a script from http://invisible-island.net/xterm/xterm.faq.html exec < /dev/tty oldstty=$(stty -g) stty raw -echo min 0 # on my system, the following line can be replaced by the line below it echo -en "\033[6n" > /dev/tty # tput u7 > /dev/tty    # when TERM=xterm (and relatives) IFS=';' read -r -d R -a pos stty $oldstty # change from one-based to zero based so they work with: tput cup $row $col row=$((${pos[0]:2} - 1))    # strip off the esc-[ col=$((${pos[1]} - 1)) 

Answer 2

You could tell read to work silently with the -s flag:

echo -en "\E[6n" read -sdR CURPOS CURPOS=${CURPOS#*[} 

And then CURPOS is equal to something like 21;3.