mySQL populated dropdown box how to update a table on form submit?

I have a dropdown box populated from a table in mySQL within a form. When I submit the form currently it records the record. I now need it to update a column within the table and change the status from "in" to "out" upon submitting the form. What code will i need to do something like this? I can get simple update mySQL working in stand alone example using the following code, but I cannot seem to get this to work with my dropdown box. Any ideas?

UPDATE table_name SET
column_name1 = value1,
column_name2 = value2,
column_name3 = value3 ...
[WHERE conditions];

PHP:

This is my current code that makes the dropdown box.

<form id="submit" name="submit" action= "ajax.php" method="post">

<select onchange="display_data(this.value);"> 

    <option>Select Laptop</option> 

    <?php 

    $query="SELECT tnumber FROM laptops WHERE status='in'";

    $result = mysql_query($query) or die(mysql_error()); 

    while(list($tnumber)=mysql_fetch_row($result)) { 

        echo "<option value=\"".$id."\">".$tnumber."</option>"; 

    } 

    ?> 

</select> 

PHP:

 

I am not getting what you are trying to ask....

 

yea, it's kindof hard to explain in words. Let me try again.

I have a input form on a page called checkout.php that uses a submit button. When you click submit it posts to a page called ajax.php, which handles the posting of a new record into a table called 'customers'. That is all working just dandy, but now I also need to update a column named 'status' in a table called 'laptops' which is a master table of laptops that are able to be checked out. Here is a screen shot to help understand what i am trying to do.

The dropdown is populated with the table 'laptops' and has a field called 'status' that i need to change from "in" to "out" when it is submitted.


Here is my ajax.php page that does the submitting of the record.

// command to start database connection



mysql_connect($mysql_hostname,$username,$password);



// Select the database

@mysql_select_db($database) or die( "Unable to select database");





	// CLIENT INFORMATION

	$tnumber      = htmlspecialchars(trim($_POST['tnumber']));

	$gid          = htmlspecialchars(trim($_POST['gid']));

	$fname        = htmlspecialchars(trim($_POST['fname']));

	$lname        = htmlspecialchars(trim($_POST['lname']));

	$ophone       = htmlspecialchars(trim($_POST['ophone']));

	$cphone       = htmlspecialchars(trim($_POST['cphone']));

	$email        = htmlspecialchars(trim($_POST['email']));

	$cdate        = htmlspecialchars(trim($_POST['cdate']));

	$rdate        = htmlspecialchars(trim($_POST['rdate']));

	$ctech        = htmlspecialchars(trim($_POST['ctech']));



    $addClient  = "INSERT INTO customers (tnumber,gid,fname,lname,ophone,cphone,email,cdate,rdate,ctech) VALUES ('$tnumber','$gid','$fname','$lname','$ophone','$cphone','$email','$cdate','$rdate','$ctech')";

    mysql_query($addClient) or die(mysql_error());



?>

PHP:

 

would it be something like this

$status = $POST['status']
$updateStatus ="UPDATE laptops SET status='out' WHERE tnumber=$tnumber" ;

mysql_query($updateStatus) or die(mysql error());

PHP:

although i never defined 'status' in the form.

 

Remove this line:

$status = $POST['status']

PHP:

Why would you need to use $_POST['status'] ? You already know the hardcoded value of status which is "out"