condition evaluating to "TRUE" - shell

Question

I am trying to check for duplicate records in my database using shell scripting.
For this, I have created a function named "check()" which echo's True or False and is stored in variable "result". But while evaluating using if statement it is always returning "True".
#redundancy check function
check() {
temp=$(grep -w -c "$1" database.dat)
echo $temp
if [ "$temp" != 0 ]
then
echo True
else
echo False
fi
}
insert() {
option="y"
while [ "$option" == "y" ]
do
echo "Rollno: \c"
read roll
result="$(check $roll)"
echo $result
if [ "$result" == "False" ]
then
echo Do something
else
echo "ERROR: Duplicate record found...\nEXITING...\n"
option="n"
fi
done
}

Answer 1

If you're using a shell that doesn't support the == extension to test, then your tests will always, unconditionally fail simply on account of invalid syntax. Use = for string comparisons to be portable to all POSIX-compliant implementations.
Moreover, there's no point to storing and then comparing the output from grep at all: Use the exit status of grep -q when your only goal is to check whether the number of matches is zero or more-than-zero; this allows grep to exit immediately when a match is seen, rather than needing to read the rest of the file.
# with -q, this emits no stdout, but exits w/ status 0 (matches exist) or 1 (otherwise)
check() { grep -q -w -e "$1" database.dat; }
insert() {
option=y
while [ "$option" = y ]; do
printf '%b\n' "Rollno: \c"
read -r roll
if check "$roll"; then
printf "ERROR: Duplicate record found...\nEXITING...\n"
option=n
else
echo "Check failed; do something"
fi
done
}