I can not identify the error that is occurring in the code below:
$bannerguia=mysqli_query($con,"SELECT id, senha FROM conta WHERE senha = '$senhalog'");
while($painel_banner=mysqli_fetch_array($bannerguia))
{
$issenha = $painel_banner['senha'];
$isid = $painel_banner['id'];
}
In SQL when I put the variable $senhalog
in quotation marks '$senhalog'
the filter does not work, if it does not put quotation marks it works, but it gives error if it does not find anything.
In all the systems I develop I put quotation marks around and everything works perfectly, I did not understand what's happening here.