Undefined index in registration form PHP - php

Question

I trying to create a simple user registration, that saves data into a database.
But I have an error I cant defeat.
Its says
Notice: Undefined index: pass_conf in C:\wamp\www\book\reg.php on
line 27
But my pass_config seems to be right.
Any ideas whats the problem ? this is the php code
/// Updating my code, this one works
<?php
$host = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "pagination";
$connection = mysql_connect($host,$dbuser,$dbpass);
$db = mysql_select_db($dbname,$connection);
/*
$name = $_POST["username"];
$pass = $_POST["password"];
$pass_conf = $_POST["pass_conf"];
$email = $_POST["email"];
$ip = $_POST["ip"]; */
$name = isset($_POST['username']) ? $_POST['username'] : 0;
$pass = isset($_POST['password']) ? $_POST['password'] : 0;
$pass_conf = isset($_POST["pass_conf"]) ? $_POST["pass_conf"] : 0; // 0 equals your default off value.
$email = isset($_POST['email']) ? $_POST['email'] : 0;
$ip = isset($_POST['ip']) ? $_POST['ip'] : 0;
if($name == false || $pass == false || $pass_conf == false || $email == false){
echo "Please fill in all the required fields.";
};
if($pass != $pass_conf){
echo "Passwords do not match.";
}else {
$connection = mysql_connect($host,$dbuser,$dbpass);
$db = mysql_select_db($dbname,$connection);
$sql = "INSERT INTO user (username,password,email,ip) VALUES ($name, $pass, $email, $ip)";
$result = mysql_query($sql);
echo "Thank you for your registration to Our Site";
};
?>
I Will add my form.php also since I got a request for it.
<?php
$IP = $_SERVER['REMOTE_ADDR'];
?>
<form name=reg action=reg.php method=post>
Username : <input type=text name=username><br>
Password : <input type=password name=password><br>
Confirm :<input type=password name=pass_conf><br>
Email : <input type=text name=email><br>
<input type=hidden name=ip value='<?php echo $IP ?>'>
<input type=submit value='Register'>
</form>
Thanks

Answer 1

instead of $x = $_POST['x'] use $x = isset($_POST['x']) ? $_POST['x'], null.
Or use:
function post($key, $default = null) {
return isset($_POST[$key]) ? $_POST[$key] : $default;
}

Answer 2

First off, good job on using E_ALL while creating your app! Your "pass_conf" value is probably a checkbox that isn't selected when submitting your form and so no value is ever sent. A simple:
$pass_conf = isset($_POST["pass_conf"]) ? $_POST["pass_conf"] : 0; // 0 equals your default off value.

Answer 3

That message means that the value you are trying to get is not present in the array. So either the value is not filled in or is not passed thru correctly (check the fields name, it has to be exactly the same).

Answer 4

Have you tried:
if (array_key_exists("pass_conf", $_POST))
$pass_conf = mysql_real_escape_string($_POST["pass_conf"]);
else
$pass_conf = null;
$sql =
"INSERT INTO events (username, password, email, ip) " .
"VALUES ('{$name}','{$pass}','{$email}','{$ip}')";
See this question that I asked for my full example and scroll down to how it was answered by user dldnh; it's similar to yours, so maybe it will help you: [link] Undefined index notice