I am getting the folowing error when this code is run:
Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource in /home/daniels/public_html/display.php on line 5
From code:
<?php
include("connect.php");
$query ="SELECT * FROM daniels_addresses";
$result= mysql_query($query);
$num=mysql_numrows($result);
?>
<HTML>
<table border="1" cellspacing="2" cellpadding="2">
<tr>
<th><font face="Arial, Helvetica, sans-serif">First Name</font></th>
<th><font face="Arial, Helvetica, sans-serif">Surname</font></th>
<th><font face="Arial, Helvetica, sans-serif">E-Mail</font></th>
</tr>
</HTML>
<?PHP
$i=0;
while ($i<$num) {
$FirstName =mysql_result($result,$i,"firstname");
$LastName =mysql_result($result,$i,"lastname");
$Email= mysql_result($result,$i,"email");
$ID=mysql_result($result,$i,"id");
?>
<HTML>
<tr>
<td><font face="Arial, Helvetica, sans-serif"><? echo "$firstname"; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? echo "$lastname"; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? echo "$Email"; ?></font></td>
</TR>
</HTML>
<?PHP
++$i;
}
?>
I apologise if it is asimple mistake but I am still learning. I would apreciate any help.


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