Limit bash while loop to 10 retries - bash

Question

How can this while loop be limited to maximum 10 retries?
#!/bin/sh
while ! test -d /somemount/share/folder
do
echo "Waiting for mount /somemount/share/folder..."
sleep 1
done

Answer 1

Keep a counter:
#!/bin/sh
while ! test -d /somemount/share/folder
do
echo "Waiting for mount /somemount/share/folder..."
((c++)) && ((c==10)) && break
sleep 1
done

Answer 2

You can also use a for loop and exit it on success:
for try in {1..10} ; do
[[ -d /somemount/share/folder ]] && break
done
The problem (which exists in the other solutions, too) is that once the loop ends, you don't know how it ended - was the directory found, or was the counter exhausted?

Answer 3

I would comment but I do not have enough points for that. I want to contribute anyway.
So this makes it work even if the while loop is nested in another loop. before the break the c variable is being reset to zero.
credits to #anubhava who came up with the original solution.
#!/bin/sh
while ! test -d /somemount/share/folder
do
echo "Waiting for mount /somemount/share/folder..."
((c++)) && ((c==10)) && c=0 && break
sleep 1
done

Answer 4

You can use until (instead of "while ! ... break), with a counter limit:
COUNT=0
ATTEMPTS=10
until [[ -d /somemount/share/folder ]] || [[ $COUNT -eq $ATTEMPTS ]]; do
echo -e "$(( COUNT++ ))... \c"
sleep 1
done
[[ $COUNT -eq $ATTEMPTS ]] && echo "Could not access mount" && (exit 1)
Notes:
Just like setting counter as variable, you can set var condition="[[ .. ]]", and use until eval $condition to make it more generic.
echo $(( COUNT++ )) increases the counter while printing.
If running inside a function, use "return 1" instead of "exit 1".