问题
i'm having weird results from php $stmt->bind_param, here is what happens... if i do this $status="something"; $var=5;
$consulta="UPDATE carrito SET status='$status' WHERE id_carrito='$var'";
and prepare and execute it works... but as soon as i do this:
$consulta="UPDATE carrito SET status=? WHERE id_carrito=?";
if($stmt=$mysqli->prepare($consulta))
{
$stmt->bind_param("si",$status,$var);
.
.
.
it stops working, vars, are ok i print them after the execute and they have the correct value actually, i don't get any error from php, query executes, it just dont save values on mysql db also, i want to mention this is not the first time i work with prepared statements i've doing this for a while now, i'm not an expert yet, but this has never happened to me before i know there is something wrong with bind_param, but i don't find any information thanks.
here is complete code
/*_____________________ DATOS DE CONEXION _________________________*/
$mysqli = new mysqli('localhost', 'root', '', 'ambarb');
if(mysqli_connect_errno())
{
echo "Connection Failed: " . mysqli_connect_errno();
exit();
}
/*_____________________ fin de los datos de conexion _______________*/
$idcarrito=$_POST["id"];
$status="cancelado";
$resultado=array();
$consulta="UPDATE carrito SET status='$status' WHERE id_carrito=$idcarrito";
if($stmt=$mysqli->prepare($consulta))
{
//$stmt->bind_param("si",$status,$idcarrito);
$stmt->execute();
if($stmt->errno==true)
{
$resultado['status']='error';
}
else
{
$resultado['status']='ok';
}
$stmt->close();
}
else
{
$resultado['status']='error al preparar la consulta PHP ERROR CODE ';
}
echo json_encode($resultado);
//echo "el ide del carrito ".$idcarrito;
$mysqli->close();
?>
that code actually works, prints $resultado['status']=ok, but i think is not the point, because as soon as i change for this $consulta="UPDATE carrito SET status=? WHERE id_carrito=?"; with respective bind_param it stops working, i mean prints $resultado['status']=ok, but doesn't make anychange at database
回答1:
You used a $var variable in first sentence, and $idcarrito in the second one. Could that be the problem ?
回答2:
I wrote a class that extends mysqli. It handles all the bind_params automatically and makes using mysqli easy. You can download the class here: better_mysqli.php
This page shows it's basic usage: basic_usage.php
This page shows detailed usage: detailed_usage.php
来源:https://stackoverflow.com/questions/17957178/php-bind-param-is-not-working