Is there some way to make su ask for the password again in case of a wrong password (just as sudo does)?
Initially I thought of this:
while ! su -c foo; do
read -n 1 -p "Retry(y/n)?" ch
if [[ "$ch" != "y" ]]; then
break
fi
done
But foo returns 1 in some cases as well, making it impossible to detect whether a wrong password was entered or if foo failed.
As a workaround, I would suggest the following:
#!/bin/bash
while ! su -c "foo; exit 0"; do
read -n 1 -p "Retry? (y/n) " ch
[[ "$ch" != "y" ]] && break # This does the same in fewer lines
done
This returns a zero value even if foo has a different exit code.
WARNING: using this workaround (which is a bit ugly), you won't be able to determine foo's exit code. I don't know if it is required for your project.
foo before exiting, so it won't "just exit".
– psimon
Jun 20 '14 at 12:34
su returns 1 on an authentication failure, so arrange for the command to avoid this error code. For example, if you don't care about conflating status values 126 and 127:
while
su -c 'foo; ret=$?; if [ "$ret" -ge 1 ] && [ "$ret" -le 126 ]; the ret=$((ret+1)); fi'
ret=$?
[ ret -ne 1 ]
do
…
done
if [ "$ret" -ge 2 ] && [ "$ret" -le 126 ]; then ret=$((ret-1)); fi
echo "foo returned $ret" # or maybe 126 if this shows 127
sudoin the first place? – terdon Jun 20 '14 at 12:36sudoon them. (I don't have control over the boxes on which this script will be run). – Jun 20 '14 at 12:39foothen? Not to keep the root shell open? In that case @psimon's answer should work for you. – terdon Jun 20 '14 at 12:42