PHP/MySQL - Update Database/Run PHP Script on Button Click - php

Question

I have a 'registration' page in PHP and I would like the script to run when an HTML button is clicked.
The PHP basically checks if all fields are filled, checks if the password and email confirmations are the same and saves to the database.
This is the code:
<?php
$Name = isset($_POST['Name']);
$Surname = isset($_POST['Surname']);
$Username = isset($_POST['Username']);
$Email = isset($_POST['Email']);
$C_Email = isset($_POST['C_Email']);
$Password = isset($_POST['password']);
$C_Password = isset($_POST['c_password']);
$SecQ = isset($_POST['SecQ']);
$SecA = isset($_POST['SecA']);
$con = mysql_connect('localhost', 'admin', 'storefile1234');
mysql_select_db ("storefile");
$check_username = mysql_query("SELECT FROM users WHERE username = '$Username'");
$check_email = mysql_query("SELECT FROM users WHERE Email = '$Email'");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
if ($Name == null || $Surname== null || $Username == null || $Password == null || $C_Password == null || $Email == null || $C_Email == null || $SecQ == null || $SecA == null ) {
echo "Missing details. Please enter all fields.";
} else {
if(mysql_num_rows($check_username) != 0 && mysql_num_rows($check_email) != 0)
{
echo "Username/Email already exists";
}
if ($Email == $C_Email && $Password == $C_Password) {
$query = "INSERT INTO users (Username, Name,Surname, Password, Email, SecQ, SecA) VALUES ('NULL', ".$Username."', ".$Name."', ".$Surname."', ".$Password."', ".$SecQ."', ".$SecA."', ".$Email.')"';
mysql_query($query) or die ('Error registering.');
echo "Greetings, ".$Name.", you have been registered. ";
} else {
echo "Error registering your account. Please try again.";
}
}
?>
Also, is it recommended?
Whenever I run this page Missing details. Please enter all fields. displays, without having entered any details.
How do you do this?

Answer 1

You tying to get values by isset($_POST['Username']); and like this functions...
But documentation says: Returns TRUE if var exists and has value other than NULL, FALSE otherwise.
So check on true, nut null. And escape your POST data after.
You can do like this:
$Name = isset($_POST['Name']) ? mysql_real_escape_string($_POST['Name']) : null;
P.S. Please again. Do not use mysql_* function. They are DEPRECATED.
Look on PDO (or mysqli_*)

Answer 2

For the issue of printing that message when you first load the page, use the array_key_exists function to test if the user has already submited something before checking if any field is null. Something like this:
if (array_key_exists('Name', $_POST) || array_key_exists('Surname', $_POST) || ... )
if ($Name == null || $Surname== null || ... )
echo "Missing details. Please enter all fields.";
Observation: you cannot use the isset function for the same purpose since, according to php documentation, it "determine if a variable is set and is not NULL"

Answer 3

You misuse isset
Try something like this:
$Name = null;
if (isset($_POST['Name'])) {
$Name = $_POST['Name'];
}
isset is only to check if a value is set.