I think I got it correct got the most part. I am having a problem with
the urlencode function I believe.
On page CustomerAddNew1.php ( page than handles form data ) im using:
$last_id = mysql_query("SELECT LAST_INSERT_ID() from customerinfo");
$last_id is now a Result Resource, not the value you're after. You must use mysql_result() or any of the mysql_fetch_*() functions to retrieve the value...
$lastid = mysql_result($last_id,0);
$last_id = urlencode ($last_id);
No need to urlencode an integer.
header("Location: UserMain.php?custid='$last_id'");
You don't put quotes around values in the URL.
and on the UserMain.php page I am using this to return the data:
$cid = "$last_id";
You called it "custid" in the URL, not "last_id". All you need here is $cid = $cust_id although that's a waste of code. What you really want is
$cid = (int)$_GET['cust_id'];
so that now you know $cid is an integer and you're not opening yourself wide open to SQL injection attacks later.
ini_set('display_errors', 1);
error_reporting(E_ALL &~ E_NOTICE);
$connect = mysql_connect("") or die ("unable to connect to database" .
mysql_error() . ""); $select = mysql_select_db("") or die ("unable to
connect to database" . mysql_error() . "SPDATA"); $result =
mysql_query("select * from customerinfo where custid='$cid'") or die
The "custid" column is an integer, right? Why are you passing it a string by putting quotes around $cid?
(mysql_error("Unable to query database")); while ($row =
mysql_fetch_array($result)) { $firstname = $row['firstname']; $lastname
= $row['lastname']; <snip>
This whole process of $var = $row['var'], $var2 = $row['var2'] is a waste of resources. Why do you need to rename the variable? You already have $row['firstname'] as a variable, just use it. If you don't understand how to use an array when printing a string, then check the manual.
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