alright, new problem, no more errors, but there's just nothing...here's the
code:
<?php
$id = rand(1,2);
$query = "SELECT name FROM mp3 WHERE songid = '" .$id."'";
$result = mysql_query($query,$connection);
while($mp3d = mysql_fetch_array($result))
{
$mp3d = mysql_fetch_array($result);
?>
<a href="http://www.fplg.net/stream.php?songid=<?php echo $id;?>"
class="hov1"><?php echo $mp3d["name"];?></a>
<?php
}
?>
</td>
</tr>
and here's what the HTML code looks like from the page when i test it out..
<a href="http://www.fplg.net/stream.php?songid=2"; class="hov1"></a>
what's going on with that...
chris
----- Original Message -----
From: James Bogosian <[EMAIL PROTECTED]>
To: Chris Cocuzzo <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Monday, July 23, 2001 7:25 PM
Subject: RE: [PHP]MySQL error, what's wrong here..
> Try:
>
> $query = "SELECT songname FROM mp3 WHERE sgid = '" .$id."'";
>
> an SQL query requires you to quote strings.
>
> james
>
> -----Original Message-----
> From: Chris Cocuzzo [mailto:[EMAIL PROTECTED]]
> Sent: Monday, July 23, 2001 6:57 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP]MySQL error, what's wrong here..
>
>
> <?php
> $id = rand(1,2);
> $query = "SELECT songname FROM mp3 WHERE sgid = " .$id;
> $result = mysql_query($query,$connection);
> $mp3d = mysql_fetch_array($result);
> ?>
>
> the server is telling me line 43(which starts with $mp3d) is not a valid
> mysql result resource. what am i doing wrong??
>
> chris
>
>
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>
>
>
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