Code:
<?php
if(isset($_POST["login"])){
function validate($input){
$input = trim(stripslashes(htmlspecialchars($input)));
return $input;
}
$formuser = validate($_POST["username"]);
$formpass = validate($_POST["password"]);
}
include("conn.php");
$sql = "SELECT username,email,password FROM login WHERE username='$formuser'";
$result = mysqli_query($conn,$sql);
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_assoc($result)){
$user = $row["username"];
$email = $row["email"];
$hashpass = $row["password"];
}
}
if( password_verify($formpass,$hashpass)){
session_start();
$_SERVER['loggeduser'] = $user;
$_SERVER['loggedemail'] = $email;
header("Location:profile.php");
}else{
$logginError = "<div class='alert alert-danger'> UserName / Password Donot match <a class='close' data-dismiss='alert'></a></div> ";
}
mysqli_close($conn);
?>
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<link rel="stylesheet" href="main.css">
</head>
<body>
<div class="box">
<form action="<?php htmlspecialchars($_SERVER["PHP_SELF"])?>" method="post">
<label> Username </label>
<input type="text" name="username" placeholder="username">
<label> Password </label>
<input type="password" name="password" placeholder="Password">
<input type="submit" name="login" value="Submit"> <input type="reset" value="Reset">
</form>
</div>
</body>
</html>Notice: Undefined variable: formuser in C:\xampp\htdocs\DIWE php\DIWE 20 Dec\form\index1.php on line 26
Notice: Undefined variable: formpass in C:\xampp\htdocs\DIWE php\DIWE 20 Dec\form\index1.php on line 44
Notice: Undefined variable: hashpass in C:\xampp\htdocs\DIWE php\DIWE 20 Dec\form\index1.php on line 44
Kiyala Error ekak enawa Form Eka wada Login wenawa But Mata dana ganna oni Ay Mehema wenne kiyala
Mama udin Isset Ekakuth gahalane tiyenne
Mekata solution eka mokakda?
Me error eka enne udin form eka yatin display wenawa
page load karaddima me variables wenama set welaa nathi nisai me error eka enne
What is the solution
