undefined variable

This eCommerce exercise is giving me an error, please see if you can help ?
Everything works fine until I do the final checkout.

These are the error lines :
Notice: Undefined variable: custid in C:\Xamp\htdocs\chapter15\checkout3.php on line 71
I thought it is already defined in line 66 ? (or is mysql_insert_id() not imported properly ?)

Notice: Undefined variable: customers_custnum in C:\Xamp\htdocs\chapter15\checkout3.php on line 90
         customers_custnum  is from the eCommerce database

Another problem is that the temporary file (createtemp.php) is not emptied and items keep piling up every time I go back to test.
Perhaps this problem will dissappear when the previous problem is solved ?
checkout3.php
Willem NorvalstudentAsked:
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Dave BaldwinFixer of ProblemsCommented:
if ($rows >= 1) then 'custid' will not be defined because it is inside that 'if' statement but you are still trying to use 'custid' in the next 'if'.  I would define a default value for 'custid' somewhere earlier in the file that will make it work the way you want.  Maybe... $custid = 0; before the first 'if'.
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Willem NorvalstudentAuthor Commented:
Thanks Dave, I'll try that
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Willem NorvalstudentAuthor Commented:
Hi Dave,
I defined a default value for  $custid = 0;  before the first 'if'.
It seemed to have worked,.

but now I'm getting this error notice :
Notice: Undefined variable: customers_custnum in C:\Xamp\htdocs\chapter15\checkout3.php on line 96

Also, the items keep piling up and it only totals the last transaction
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Willem NorvalstudentAuthor Commented:
Hi Dave,
I inserted the default value for $custid = 0;     before the first 'if'.   It worked.

However, my previous  carts are not emptied and there's a problem with the total - it only gives the total of the last cart.
And it's now giving this error  :
Notice: Undefined variable: customers_custnum in C:\Xamp\htdocs\chapter15\checkout3.php on line 96
(which is '$total', )
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Dave BaldwinFixer of ProblemsCommented:
It's the same problem but with a different variable.  You are defining it inside an 'if' and then trying to use it whether it is defined or not.  You have to define a value for a variable before you can use it.  I also do not see any purpose for using $customers_custnum = $custid;.  Just use $custid.

Most of my PHP pages start out with a long list of variable definitions to make sure things like this do not happen.  In my code, lines like this...
$firstname = $_POST['firstname'];

Open in new window

are done like this:
if (!isset($_POST['firstname']))  $firstname = ''; else $firstname = $_POST['firstname'];

Open in new window

This makes sure that the POST data always has a defined value.
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