bash - How to use goto statement in shell script -

i beginner in shell script. don't have idea how use goto statement. using following code.

start: echo "main menu" echo "1 copy" echo "2 exit" read num case $num in "1") echo "copynum" goto start:; "2")          echo "haiiii"; goto start: *) echo "ssss"; esac 

as others have noted, there's no goto in bash (or other posix-like shells) - other, more flexible flow-control constructs take place.
heading compound commands in man bash.

in case, select command right choice. since how use may not obvious, here's started:

#!/usr/bin/env bash  echo "main menu"  # define choices present user, # presented line line, prefixed sequential number # (e.g., '1) copy', ...) choices=( 'copy' 'exit' )  # present choices. # user chooses entering *number* before desired choice. select choice in "${choices[@]}";    # if invalid number chosen, $choice empty.   # report error , prompt again.   [[ -n $choice ]] || { echo "invalid choice." >&2; continue; }    # examine choice.   # note choice string itself, not number   # reported in $choice.   case $choice in     copy)       echo "copying..."       # set flag here, or call function, ...       ;;     exit)       echo "exiting. "       exit 0   esac    # getting here means valid choice made,   # break out of select statement , continue below,   # if desired.   # note without explicit break (or exit) statement,    # bash continue prompt.   break  done