1. Check the manual on use of $_POST and notch down your error reporting
level. As these vars have not been defined PHP is throwing an error.
2. While you are at it, and assuming that your variables are all received
correctly, why not simplify your sql?
insert into test_scores set field_one =
Chris,
It appears that the indices called out in the lines of error code are not
being transferred from the Insert_Table_Form.html page and that is causing
the query to ultimately fail. Almost self-explanatory due to PHP's good
error reporting.
First make certain that the names of the form contro
It does not appear that you have posted the code from the add.php page in
your original post (below). From the error messages you provided, that seems
to be where the error is happening. You'll need to post the add.php code and
identify line #23 for us to be able to help.
> -Original Message--
Your best bet is to do some isset() calls:
eg
if(isset($row['imageurl']))
{
$image=$row['imageurl'];
}
else
{
/*maybe report the problems or handle it like this...*/
$image=$default_image_url;
/*where you have already set up $default_image_url to point somewhere
sensible*/
}
Cheers
--
Phil
I don't think dropping the quotes is the best solution.
One might consider dropping the warning level in php.ini, if it is a
problem, or use numbered array like mysql_fetch_row() returns.
-Joe
""Henning Kilset Pedersen"" <[EMAIL PROTECTED]> wrote in message
000f01c0d251$59b79000$a26547c1@peders
Hi!
Drop the quotes around your array keys. It should be $row[imageurl] for
example.
--
Henning Kilset Pedersen
Anarchy Online Server Operations
Oracle, PHP, e-Commerce etc.
Funcom Oslo AS
-Original Message-
From: Petra [mailto:[EMAIL PROTECTED]]
Sent: 1. mai 2001 03:29
To: [EM
