Hi John,
You're trying to use a variable that is not set. Try using the isset
function... Like this:
if (isset($COCOD))
$result = mysql_query("SELECT * FROM tblAA_MIKER_COMPDAT WHERE
COCOD=$COCOD",$db);
Hope this helps...
Later,
Jorge
[EMAIL PROTECTED]
----- Original Message -----
From: John Halladay <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, February 14, 2001 12:51 PM
Subject: [PHP-DB] Undefined Variable and argument not valid . . .
> Could someone please give me some direction on what is wrong with what I
am
> trying to do below. I've checked it over but still get the same errors.
> What I'm trying to do is have a list of links that display information
when
> you click them.
>
> I'm getting the following error at the top of the first page where all the
> links are displayed.
> Warning: Undefined variable: COCOD in C:\Inetpub\wwwroot/maybe.php on line
> 11
>
> I'm getting this error when I click on one of the links and there is no
info
> displayed under the three listing below.
> Warning: Supplied argument is not a valid MySQL result resource in
> C:\Inetpub\wwwroot/maybe.php on line 15
> Company Name:
> Contact:
> Phone Number:
>
>
> <html>
>
> <body>
>
> <?php
>
> $db = mysql_connect("localhost", "root", "*******");
>
> mysql_select_db("schedules",$db);
>
> if ($COCOD) {
>
> $result = mysql_query("SELECT * FROM tblAA_MIKER_COMPDAT WHERE
> COCOD=$COCOD",$db);
>
> $myrow = mysql_fetch_array($result);
>
> printf("Company Name: %s\n<br>", $myrow["CONAM"]);
>
> printf("Contact: %s\n<br>", $myrow["Contact"]);
>
> printf("Phone Number: %s\n<br>", $myrow["PhoneNumber"]);
>
> } else {
>
> $result = mysql_query("SELECT * FROM tblAA_MIKER_COMPDAT",$db);
>
> if ($myrow = mysql_fetch_array($result)) {
>
> do {
>
> printf("<a href=\"%s?COCOD=%s\">%s</a><br>\n", $PHP_SELF, $myrow["COCOD"],
> $myrow["CONAM"]);
>
> } while ($myrow = mysql_fetch_array($result));
>
> } else {
>
> echo "Sorry, no records were found!";
>
> }
>
> }
>
> ?>
>
> </body>
>
> </html>
>
> Thanks.
>
> John Halladay
>
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