bash script with parameters in a while loop [duplicate] - bash

Question

This question already has answers here:
Read user input inside a loop
(6 answers)
Closed 4 years ago.
the script is:
#!/bin/bash
SCRIPTLOG=~/tmp/logfile
[ ! -d ~/tmp ] && mkdir ~/tmp
while getopts ":u:g:y" options; do
case $options in
u )
user=$OPTARG;;
g )
group=$OPTARG;;
* )
echo "wrong option: ${OPTARG}"
exit 1;;
esac
done
set -e
add_etx_members ()
{
user="${1}"
group="${2}"
set -u
echo "Users to add to ${group}:"
echo
echo "--- BEGIN ---"
printf "%-10s\n" $user
echo "--- END ---"
echo
while true; do
read -r -p "Continue? [y/n]: " REPLY
case $REPLY in
[yY]) echo;break ;;
[nNqQ]) echo;exit ;;
*) printf "\033[31m%s\033[0m\n" " invalid input: ${REPLY}"
esac
done
echo "here commands"
}
add_etx_members "${user}" "${group}" 2>&1 | tee -a ${SCRIPTLOG}
when I run single execution from command line it works (reaches "echo here commands"):
$ myscpt1 -u test -g ABC
Users to add to ABC:
--- BEGIN ---
test
--- END ---
Continue? [y/n]: y
here commands
$
but when I run it in while loop, it fails:
$ echo -e "ABC\nSDE"|while read a;do myscpt1 -u test -g ${a};done
Users to add to ABC:
--- BEGIN ---
test
--- END ---
invalid input: SDE
$
Although, the commands are the same as from single run:
$ echo -e "ABC\nSDE"|while read a;do echo myscpt1 -u test -g ${a};done
myscpt1 -u test -g ABC
myscpt1 -u test -g SDE

Answer 1

because the read inside the script is consuming the same input (the standard input is inherited from caller) as the while read calling the script.
may be resolved redirecting input inside script
exec < /dev/tty
or just for the while read
while ...;
done < /dev/tty
Note that while true could be changed by while read
or from outside depending on what you need
echo -e "ABC\nSDE"|while read a;do myscpt1 -u test -g ${a} < /dev/null;done