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 fo

Answer 1

In the interests of portability I've had a go at making a vanilla POSIX-compatible version that will run in shells like dash:

#!/bin/sh

exec < /dev/tty
oldstty=$(stty -g)
stty raw -echo min 0
tput u7 > /dev/tty
sleep 1
IFS=';' read -r row col
stty $oldstty

row=$(expr $(expr substr $row 3 99) - 1)        # Strip leading escape off
col=$(expr ${col%R} - 1)                        # Strip trailing 'R' off

echo $col,$row

...but I can't seem to find a viable alternative for bash's 'read -d'. Without the sleep, the script misses the return output entirely...