REALLY need some help. Passing Variables to Database?

persianboi

I'm newbie to PHP and learning things on my own. Anyways i have a project to finish by tomorrow and it's really not going well. I can try to simplify this as much as possible so u guys understand.

Introduction:

Reservation form online. so say we have a reservation.inc that includes the form and it has 1 field at this point cause if I figure out the problem with one of them, the rest is ok. 2- there is also login2.php that recieves data, checked if its ok, if so, writes into a database.

<form action="Login2.php?do=new" method="post" >
		<table border="0" width="420">
		<?PHP
		if (isset($message_new))
		echo "<tr><td colspan='2'><b>$message_new</b>
		</td></tr>";
		?>
		<tr>
		<td class="subtitleb">* <span class="text">Arrival date:</span>
		  <input type="text" name="nnights" size="10" maxlength="4" value="<?php echo @$ndyear ?>">

As u see the name is nnights.

2- Login page

<?php 

session_start();
include("info.inc");

switch (@$_GET['do'])
{
	case "new":

foreach($_POST as $field => $value)
/* the code in the middle here is not important right now because its just checking against pattern */

/* My problem starts here */

$yearstring = '$ndyear';
				$monthstring = '$ndmonth';
				$daystring ='$ndday';
				$newdate = $yearstring.$monthstring.$daystring;

			$sql = "INSERT INTO reservations (numNites)
					VALUES ( '$nnights')";
			mysql_query($sql);

Notice: Undefined variable: nnights in C:\Program Files\Apache Group\Apache2\htdocs\Pelagos\Login2.php

and I get this same problem with every other variable. But I know infact they are pass along because when I test the Pattern check section it works like a charm.

Oh and

default:
		include("reservations.inc");

is included in the bttm

Can someone PLZ help me out. I really need to finish this project. I'd GREATLY appriciate some help

herve

You shouln't use ' character around a PHP variable but " otherwise the variable will not
be interpreted and replace by its content.

$myvar = 1;
$myvar2 = $myvar;

is OK

$myvar = 1;
$myvar2 = "$myvar";

is useless but OK

$myvar = 1;
$myvar2 = '$myvar';

won't work.

Also if PHP tell you that a variable is not defined, it mean there's no content inside
that variable. In the code you've showed us, there's no

$nnights = " somevalue " ;

nowhere.

So on the line

$sql = "INSERT INTO reservations (numNites)  VALUES ( '$nnights')";

there's a logical error if there's no content in $nnights