Good morning!
I'm new in PHP. i'm trying to make work this scrpt but is showing me this problem. It's a forma that modifies some records in a mysql database. The codes gets the data but it shows me that mistake and when i run modify it shows me several problems....
Warning: mysqli_query() expects parameter 2 to be string, object given
in C:wampwwwCTEformedicion.php on line 15
I still don't know how to fix it. I really appreaciate your help.
Thanks!
<body>
<?php
include "conexiondb.php";
if(!isset($_POST['submit'])){
//$busqueda=$con->query(
$muestra=$con->query("SELECT * FROM clientes C INNER JOIN producto P ON C.serial = P.serial WHERE P.serial = $_GET[serial]");
//mysqli_query($con,$sql);
mysqli_query($con,$muestra);
$person=$muestra->fetch_array();
}
?>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
cliente<input type = "text" name="inputcliente" value = "<?php echo $person['cliente']; ?>" /><br/>
cedula <input type = "text" name="inputcedula" value = "<?php echo $person['cedula']; ?>" /><br/>
<input type="hidden" name="serial" value="<?php echo $_GET['serial'];?>"/>
<input type = "submit" name = "submit" value= "Modificar"/>
</form>
<?php
if (isset($_POST ['submit'])){
$u = "UPDATE cliente SET'cliente'='$_POST[inputcliente]','cedula' = '$_POST[inputcedula]' WHERE serial=$_POST[serial]";
mysqli_query($con,$u);
echo "El usuario ha sido modificado";
header ("Location:busca.php");
} else {
}
?>
</body>
</html>
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