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

My (two) version of same...

As a function, setting specific variable, using ncurses's user defined comands:

getCPos () { 
    local v=() t=$(stty -g)
    stty -echo
    tput u7
    IFS='[;' read -rd R -a v
    stty $t
    CPos=(${v[@]:1})
}

Than now:

getCPos 
echo $CPos
21
echo ${CPos[1]}
1
echo ${CPos[@]}
21 1

declare -p CPos
declare -a CPos=([0]="48" [1]="1")

Nota: I use ncurses command: tput u7 at line #4 in the hope this will stay more portable than using VT220 string by command: printf "\033[6n"... Not sure: anyway this will work with any of them:

getCPos () { 
    local v=() t=$(stty -g)
    stty -echo
    printf "\033[6n"
    IFS='[;' read -ra v -d R
    stty $t
    CPos=(${v[@]:1})
}

will work exactly same, while under VT220 compatible TERM.

More info

You may found some doc there:

VT220 Programmer Reference Manual - Chapter 4

4.17.2 Device Status Report (DSR)

...

Host to VT220 (Req 4 cur pos)  CSI 6 n       "Please report your cursor position using a CPR (not DSR) control sequence."
  
VT220 to host (CPR response)   CSI Pv; Ph R  "My cursor is positioned at _____ (Pv); _____ (Ph)."
                                              Pv =  vertical position (row)
                                              Ph =  horizontal position (column)