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

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.