This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
The 3 different equals
(5 answers)
Closed 2 years ago.
I have this code in my login.hmtl in s section with ajax method:
$.ajax ({
type:'post',
url:'Controladores/login.php',
data:{do_login:"do_login", nick:nick, pass:pass },
success:function(response) {
alert($.trim(response));
if ($.trim(response) ==='Correcto'){
window.location.href="intranet.php?ruta=inicio";
return false;
}
else{
alert('Datos incorrectos. Volver a intentar.');
}
}
});
This redirects me to login.php where I connect to the database and then check if the username and password are correct. If they are, they return an answer Correct or Incorrect.
Here is the main part where I start my session, and also if username and password match, set the username 'NICK' to the session.
session_start();
$db = new Conexion();
$conn = $db->connect( '00.00.00.00', 'Seguridad');
$param = [$nick, $pass];
$query = "exec Seguridad.dbo.autenticacion ?, ?";
$result= $db->rows($conn, $query, $param);
$obj = json_decode($result);
//echo $obj[0]->resultado;
if( $obj[0]->resultado ='Correcto' ){
$_SESSION['nick']= $nick;
$_SESSION['nombreusuario'] = $obj[0]->nombreusuario;
$_SESSION['agencia'] = $obj[0]->CodigoAgencia;
echo ("Correcto");
}else{
if ($obj[0]->resultado =='Incorrecto'){
echo ("Incorrecto");
}
}
Then, In my home page called inicio.php I called again the session but the message is:
Notice: Undefined index: nick in C:\xampp\htdocs\intranetv3\Vistas\modulos\inicio.php on line 3
<?php
session_start();
$userName = $_SESSION['nick'];
echo "$userName";
?>
I have read other article but have not found anything.
I checked if the answer Correct or Incorrect returns the answer and it returns.
I printed the user name and password. It worked.
I also printed the $_SESSION['nick'], I see it has the username.
Apparently when I go to my inicio.hmtl the session just unssets or something, I have no idea what might be wrong. Please HELP!!
Related
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
How to use store and use session variables across pages?
(8 answers)
Closed 2 years ago.
I'm from the Netherlands and my English isn't perfect, but I will try my best.
My problem is as followed:
I've made a local website with Wampserver and everything worked fine on it, but now I've bought hosting and my SESSION variable doesn't work anymore.
The code of the function:
public function login($uname,$umail,$upass)
{
try
{
$stmt = $this->db->prepare("SELECT * FROM users WHERE user_name=:uname OR user_email=:umail LIMIT 1");
$stmt->execute(array(':uname'=>$uname, ':umail'=>$umail));
$userRow=$stmt->fetch(PDO::FETCH_ASSOC);
if($stmt->rowCount() > 0)
{
if(password_verify($upass, $userRow['user_pass']))
{
$_SESSION["user_session"] = $userRow['user_id'];
return true;
}
else
{
return false;
}
}
}
catch(PDOException $e)
{
echo $e->getMessage();
}
}
Code of the page where I try to check if someone is logged in:
if(!$user->is_loggedin())
{
$user->redirect('login.php');
}
$user_id = $_SESSION['user_session'];
And then the error I'm getting:
Undefined index: user_session in /mnt/web515/b0/72/510494272/htdocs/admin/index.php on line 8
Can someone please help me figure out why user_session is undefined now but not when I locally run the website?
Thanks in advance!
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Reference - What does this error mean in PHP?
(38 answers)
Closed 3 years ago.
I am connected to my local host and am trying to use a GET method on line $. I am getting the Notice: Undefined index: deleteid in C:\xampp\htdocs\webd153\delete.php on line 4.
<?php
include 'connection.php';
$deleteid = $_GET['deleteid'];
if (isset($deleteid)) {
$deletesql = $dbh->prepare("DELETE FROM users WHERE id = '$deleteid'");
$deletesql->execure();
echo "record has been deleted!<br>";
I am trying to delete names that I have entered in my databases using a form that is connected from my local host to myphpadmin database.
Right way is:
<?php
include 'connection.php';
if(isset($_GET['deleteid']) {
$deleteid = $_GET['deleteid'];
$deletesql = $dbh->prepare("DELETE FROM users WHERE id = '$deleteid'");
$deletesql->execute();
echo "record has been deleted!<br>";
}
But this is VERY INSECURE! When I send request with URL ending ?deleteid=1'+OR+1=1+OR+id=', your database will be deleted all rows. I suggest to change query building as:
$deletesql = $dbh->prepare("DELETE FROM users WHERE id = (?)");
$deletesql->bind_param('i', $deleteid);
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 5 years ago.
I'm new in programming, and i don't know why it doesn't work. I use the same code for my other input form, and that code work just fine. But, for this form, it's not. The form is pretty much same, so, that's why i totally do not understand why it's not working.
This is the php code where the message error come from. I wish you can help me. Thank you so much.
<?php
session_start();
require 'db.php';
$kodeunit = $_SESSION["KodeUnit"];
$namaunit = $_POST['namaunit'];
$alamat = $_POST['alamat'];
$pimpinanunit = $_POST['pimpinanunit'];
$kuasaanggaran = $_POST['kuasaanggaran'];
$pembuatkomitmen = $_POST['pembuatkomitmen'];
$penanggungjawab = $_POST['penanggungjawab'];
$sql = "UPDATE unit_organisasi SET Nama_Unit = '$namaunit', Pimpinan_Unit = '$pimpinanunit', Alamat = '$alamat', Kuasa_Anggaran = '$kuasaanggaran', Pembuat_Komitment = '$pembuatkomitmen', Penanggungjawab = '$penanggungjawab' WHERE Kode_Unit = '$kodeunit'";
if((!strlen(trim($namaunit))) || (!strlen(trim($alamat))) || (!strlen(trim($pimpinanunit))) || (!strlen(trim($kuasaanggaran))) || (!strlen(trim($pembuatkomitmen))) || (!strlen(trim($penanggungjawab)))){
echo "<script>alert('Data Belum Lengkap!')</script>";
header ('Location:inpudataunit.php');
}
else{
$result = $conn->query($sql);
if($result === TRUE){
echo "Berhasil diinput";
}
}
?>
Seems like your form has no input field named namaunit . Check your form input name.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 5 years ago.
login code
<?php
include("koneksi.php");
$email=$_POST['email'];
$password=md5($_POST['password']);
$q="SELECT * FROM `user` WHERE `user`.`email`='$email' AND `user`.`password`='$password'";
$qe=mysql_query($q);
while ($de=mysql_fetch_array($qe)) {
$id_user=$de['id_user'];
}
if (mysql_num_rows($qe)>0) {
session_start();
$_SESSION['x']=$id_users;
header('location:home.php');
exit;
} else{
header('location:login_user.php');
exit;
}
?>
after login i wanna show or echo the username with this code
<?php
session_start();
$id_user=$_SESSION['x'];
$q="SELECT * FROM `user` WHERE `user`.`id_user`='$id_user'";
$qe=mysql_query($q);
$de=mysql_fetch_array($qe);
$username=$de['username'];
echo "
<li>$username</li>
";
?>
and the problem is the username doesn't show..
whats wrong.. help me ..
You have a typo in your login search query. Change to $_SESSION['x'] = $id_user;
Not $id_users
A piece of advice, use an IDE such as netbeans or eclipse or phpstorm. They help in identifying unused variables and other minor syntax errors.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
Hello i'm new to php and i wanted to create a admin panel and i got an error with my register page.
This is the error :
Notice: Undefined index: username in C:\wamp\www\Website2 - Copy\register.php on line 18
if (isset($_POST['submit'])){
$username = $_POST['username']; // error here
$password = md5($_POST['password']);
if(empty($username)or empty($password)){
echo"There is an empty space";
}else{
mysql_query("INSERT INTO users VALUES('', '$username', '$password')");
}
}
This happens because the empty function checks if the variables $username and $password has no value or set equal to zero. (0, null, false,''). As the empty function does not check if the variable exists, php throws this error.
In this case it would be correct to use the isset function of php.
http://br1.php.net/isset