Hi John,
Considering that I don't have MySql installed here and therefore can't test
your code, I'm going to guess that the msql_fetch_array is not storing the
array as per the indexes that you want. Try using MYSQL_ASSOC or MYSQL_BOTH
as the second parameter. MYSQL_ASSOC causes the elements to be indexed by
field names and MYSQL_BOTH FORCES the indexes to be stored both ways. It may
be possible that your default is setup to MYSQL_NUM and that would be why
you're getting an error.
Try:
$myrow = mysql_fetch_array($result,MYSQL_ASSOC);
Hope this helps.
Later,
Jorge
[EMAIL PROTECTED]
----- Original Message -----
From: John Halladay <[EMAIL PROTECTED]>
To: 'Jorge Santos' <[EMAIL PROTECTED]>
Sent: Wednesday, February 14, 2001 1:11 PM
Subject: RE: [PHP-DB] Undefined Variable and argument not valid . . .
> Great! That fixed the first error, but I'm still getting the other error
on
> the second page stating:
>
> Warning: Supplied argument is not a valid MySQL result resource in
> C:\Inetpub\wwwroot/maybe.php on line 15
> Company Name:
> Contact:
> Phone Number:
>
> Here is line 15:
>
> $myrow = mysql_fetch_array($result);
>
> Any ideas? Thanks again for your help.
>
> John
>
> -----Original Message-----
> From: Jorge Santos [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, February 14, 2001 11:05 AM
> To: John Halladay
> Cc: PHP-DB
> Subject: Re: [PHP-DB] Undefined Variable and argument not valid . . .
>
>
> 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
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
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> >
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