I am developing a website and I have insert query but it doesn't work(it isn't insert the data in my database) even though there are no errors appeared ,I am using localhost XAMMP and i put the connection code in another page this is my connection code a and my insert code.I though the problem with the connection but it print the statement
<?php
$dbhost="localhost";
$dbuser="root";
$dbpass="";
$db="Driving1";
$conn = new mysqli($dbhost, $dbuser, $dbpass, $db);
if($conn->connect_error){
echo "connection was failed ";
}
else{
echo "connection made";
}
?>
<?php
session_start();
$username = "";
$email = "";
$number ="";
$password1= "";
$password2= "";
$errors = array();
include 'conn.php';
if (isset($_POST['submit'])) {
$username = mysqli_real_escape_string($conn, $_POST['username']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$password1 = mysqli_real_escape_string($conn, $_POST['pass1']);
$password2 = mysqli_real_escape_string($conn, $_POST['pass2']);
$number = mysqli_real_escape_string($conn, $_POST['mobile']);
if ($password1 != $password2) {
array_push($errors, "The two passwords do not match");
}
$user_check_query = "SELECT * FROM Trainee WHERE name='$username' ";
$result = mysqli_query($conn, $user_check_query);
$user = mysqli_fetch_assoc($result);
if ($user) {
if ($user['name'] === $username) {
array_push($errors, "Username already exists");
}
}
if (count($errors) == 0) {
$password = md5($password1);
$query = "INSERT INTO Trainee (email, name, password,number)
VALUES('$email','$username','$password','$number')";
$result1=mysqli_query($conn, $query);
$_SESSION['username'] = $username;
$_SESSION['success'] = "You are now logged in";
//header('location: index.php');
} }
?>
Aucun commentaire:
Enregistrer un commentaire