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I cannot identify the error that is happening 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 $senhalog
in quotes '$senhalog'
does not work filter, if I do not quote works, but gives error if not locate anything.
In all the systems I develop I put in quotes and everything works perfectly, I do not understand what is happening here.
Give an echo in the query to see how the return is coming.
– David Dias
Check whether the variable
$senhalog
is at the correct value, if you are testing sql directly in the terminal or phpmyadmin or Workbench or something, the error may be in sql– Costamilam