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PHP Error
Posted by: Marc Birch
Date: April 19, 2009 12:25PM

Hello,
I'm new to PHP and mySQL and I've been following a tutorial from phpf1.com (url: http://www.phpf1.com/tutorial/php-form.html?page=1). I'm trying to make a simple form that's validated and then stored in a database. Though I keep encountering errors in the coding (I noticed some errors in the HTML, as I'm quite knowledgable with it) and I was wondering if anyone could could take a look at it?

<?php
// function to display form
function showForm($errorName = false, $errorEmail = false, $errorMesg = false) {
	if ($errorName) $errorTextName = "Please enter your name!";
	if ($errorEmail) $errorTextEmail = "Please enter a valid email address!";
	if ($errorMesg) $errorTextMesg = "Please leave a longer message!";

	echo '<form action="form.php" method="POST"><table>';
	
	// display name field and error if required
	echo '<tr><td>Name</td><td><input type="text" name="name"></td></tr>';
	if ($errorName) echo "<tr><td colspan='2'>$errorTextName</td></tr>";
	
	// display email field and error if required
	echo '<tr><td>Email</td><td><input type="text" name="email"></td></tr>';
	if ($errorEmail) echo "<tr><td colspan='2'>$errorTextEmail</td></tr>";

	// display message field and error if required
	echo '<tr><td>Message</td><td><textarea name="mesg"></textarea></td></tr>';
	if ($errorMesg) echo "<tr><td colspan='2'>$errorTextMesg</td></tr>";
	
	echo '<tr><td><input type="submit" name="SubmitForm" value="Send"></td></tr>';
	echo '</form>';

	if (!isset($_POST['SubmitForm'])) {
		showForm();
	} else {
		// init error variables
		$errorName = false;
		$errorEmail = false;
		$errorMesg = false;

		$name = isset($_POST['name']) ? trim($_POST['name']) : '';
		$email = isset($_POST['email']) ? trim($_POST['email']) : '';
		$mesg = isset($_POST['mesg']) ? trim($_POST['mesg']) : '';

		if (!eregi("^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$";
		if (strlen($name)<3) $errorName = true;
		if (strlen($mesg)<10) $ errorMesg = true; 

		// display the form again as there was an error
		if ($errorName || $errorEmail || $errorMesg) {
			showForm($errorName, $errorEmail, $errorMesg);
		} else {
			echo 'Submission was successful!';
		}
	} // end else condition
} // end function 
?>

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PHP Error
April 19, 2009 12:25PM
April 19, 2009 04:07PM


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