Checkbox value not displaying - php

Question

The form inputs aren't displaying on the form.php page and negates my form validation. The error says undefined variable for all my variables on form.php. Please tell me what I have to edit in my code to make it display the form inputs on form.php. It works when I use it on the same page but I would rather it display on another page.
EDIT
Thanks so far but I can't get the value of the checkbox, the recipient(Administrator or Content Editor), to display it displays "Array" or "A".
contact.php
<?php
$errnam = "";
$errmail = "";
$errsub = "";
$errrec = "";
$hasErrors = false;
if(isset ($_POST['submitted'])){
$name = $_POST['name'];
$email = $_POST['email'];
$subject = $_POST['subject'];
$recipient = $_POST['recipient'];
$message = $_POST['message'];
if(preg_match("/^[\w\-'\s]/", $_POST['name'])){
$name = $_POST['name'];
}
else{
$errnam ='<strong>Please enter a name.</strong>';
$hasErrors = true;
}
if (preg_match("/^[\w.-_]+#[\w.-]+[A-Za-z]{2,6}$/i", $email)){
$email = $_POST['email'];
}
else{
$errmail = '<strong>Please enter a valid email.</strong>';
$hasErrors = true;
}
if(preg_match("/^[\w\-'\s]/", $_POST['subject'])){
$subject = $_POST['subject'];
}
else{
$errsub = "<strong>Please enter a subject.</strong>";
$hasErrors = true;
}
if (!empty($_POST['recipient'])) {
for ($i=0; $i < count($_POST['recipient']);$i++) {
$recipient = $_POST['recipient'];
}
}else{
$errrec = "<strong>Please select a recipient</strong>";
$hasErrors = true;
}
$message = $_POST['message'];
}
if ($hasErrors){
echo "<strong>Error! Please fix the errors as stated.</strong>";
}else{
header("Location: form.php?name=".$name."&email=".$email."&subject=".$subject. "&recipient=".$recipient. "&message=".$message);
exit();
}
?>
form.php
<?php
$name = $_GET['name'];
$email = $_GET['email'];
$subject = $_GET['subject'];
$recipient = $_GET['recipient'];
$message = $_GET['message'];
echo "<h2>Thank You</h2>";
echo "<p>Thank you for your submission. Here is a copy of the details that you have sent.</p>";
echo "<strong>Your Name:</strong> ".$name. "<br />";
echo "<strong>Your Email:</strong> ".$email. "<br />";
echo "<strong>Subject:</strong> ".$subject. "<br />";
echo "<strong>Recipient:</strong>" .$recipient. "<br />";
echo "<strong>Message:</strong> <br /> " .$message;
?>

Answer 1

If you would like to transfer the data from contact.php to form.php you should use something like this:
contact.php
$data = urlencode(
serialize(
array(
"name" => $name,
"email" => $email,
"subject" => $subject,
"message" => $message)
));
header('Location: form.php?data=' . $data);
form.php
$data = unserialize(urldecode($_GET['data']));
$name = $data["name"];
$email = $data["email"];
$subject = $data["subject"];
$message = $data["message"];
This serializes the array of data from contact.php then URL encodes it and sends it as a GET variable to form.php. After, form.php URL decodes and unserializes the data for use.

Answer 2

The problem is when you header("Location:") to form.php, all the POST values are lost. You have to either resend them with the header, or modify them into GET and retrieve them again. It should be more efficient to have them both (contact.php AND form.php) in one page. That way, the form data only has to be sent once.
You could probably just send the POST values as GET over to form.php like this.
contact.php:
header("Location: form.php?name=".$name."&email=".$email."&subject=".$subject."&message=".$message);
form.php (to retrieve the values):
$name = $_GET['name'];
$email = $_GET['email'];
$message = $_GET['message'];
$subject = $_GET['subject'];

Answer 3

If you want to display form elements then you have to use this approach.
<form method="POST" action="contact.php">
Email<input type="text" name="email">
.......
.......
.......
// All elements
</form>
This may help you.

Answer 4

Give action in your form in contact.php
<form action="form.php">