Figured it was something simple. Part of the learning process, I suppose.
Thanks for the help!

Jason Soza

-----Original Message-----
From: Dan Koken [mailto:[EMAIL PROTECTED]]
Sent: Saturday, June 15, 2002 11:31 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Re: Code Structure/Errors


Can simply set $i before the while.

I assume the $last_name is coming from the DB.

Not exactly sure what you want, but hope this helps.
good luck... Dan.
----------------------
$i
        = 0;
while ($row = mysql_fetch_array($result)) {
        extract($row);
if (($i % 5) == 0) print "<tr>\n";
        $i++;
        @$last_name .= '';
        if($last_name <> '')
        <print stuff here>
         else
        <print other stuff>
...
if (($i % 5) == 0) print "</tr>\n";
}




Jason Soza wrote:

> I'm just curious if there's a way to restructure my code so as to avoid
> getting "Undefined Variable" errors. I keep getting them and I know
they're
> nothing to worry about for the most part, I'd like to get rid of them
> without turning off error reporting if possible. In the following code, I
> get an undefined variable error for $i and for $last_name, is there
anything
> I can do to actually define them? $last_name is a variable produced by my
> MySQL query, $i is just a counter:
>
> while ($row = mysql_fetch_array($result)) {
>       extract($row);
>       $i++;
>       if($i=="1") {
>               print "<tr>\n";
>                       }
>
>                       if($last_name) {
>                               <print stuff here>
>                               } else {
>                               <print other stuff>
>                               }
> ...
>                               if ($i=="5") {
>                                            print "</tr>\n";
>                                            $i=0;
>                                            }
>                               }
>
> Thanks!
>
> Jason Soza


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