I'm creating a program connected to a database where I have a login and then two other pages. When I click the button on the 'pagina 1' it goes to the 'pagina 2', in this page I have a button and I want to put an image or an icon on it. Is there a way to do it?
<html>
<head>
<title> Agenzia Viaggi </title>
</head>
<body>
<?php
$servername = "localhost";
session_start();
if(!(isset($_POST["pagina"]))){
echo "
<form action='Login.php' method='post'>
User:<input type='text' name ='utente' /> <br/>
Password: <input type='password' name ='pwd'/><br/>
<input type ='submit' value='Accedi'/>
<input type='hidden' name='pagina' value='1' />
</form>";
}else if($_POST["pagina"]==1){
$utente = $_POST["utente"];
$password = $_POST["pwd"];
$conn = new mysqli($servername, $utente, $password, 'agenziaviaggi');
if($conn->connect_errno){
echo "Connessione impossibile: ".$conn->connect_error;
exit;
}
$_SESSION["utente"] = $utente;
$_SESSION["pwd"] = $password;
echo "Aeroporto ";
echo "<form action ='Login.php' method = 'post'>
<input type = 'submit' name = 'aeroporto' value = 'Visualizza Aeroporto'>
<input type = 'hidden' name = 'pagina' value = '2'/>
</form>";
}
else if($_POST["pagina"]==2){
$utente = $_SESSION["utente"];
$password = $_SESSION["pwd"];
$conn = new mysqli($servername, $utente, $password, 'agenziaviaggi');
if($conn->connect_errno){
echo "Connessione impossibile: ".$conn->connect_error;
exit;
}
echo 'Aeroporto<br/>' ;
$sql = "SELECT Nome, Città, Servizi, Parcheggi, Telefono FROM aeroporto";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<br> Nome: ". $row["Nome"]. " - Città: ". $row["Città"]. " - Servizi: ". $row["Servizi"]. " - Parcheggi: ". $row["Parcheggi"]. " - Telefono: ". $row["Telefono"]. "<br>";
}
} else {
echo "0 results";
}
echo "<form action ='Login.php' method = 'post'>
<input type = 'submit' value = 'Go back'>
<input type = 'hidden' name = 'pagina' value = '1'/>
</form>";
}
?>
</body></html>
