This problem is similar to Problem W$4\mathrm{.}1.,$ except that the hash value is not calculated

inside the SQLstatement; it is calculated in the PHP code using PHP$$s hash$()$function$.$

Does this modified program have a SQL injection problem?

$hashed$\_$eid $=$ hash$($sha$256,$ $eid$)$;

$hashed$\_$passwd $=$ hash$($sha$256,$ $passwd$)$;

$sql $=$ "SELECT $*$ FROM employee

WHERE eid$=$$hashed$\_$eid$$ and password$=$$hashed$\_$passwd$"$;