What I'm attempting is to have database driven links on my page. It is all ok except one thing and that is this error when my page comes up in the browser.
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/admin/public_html/index.php on line 114
Here is my code.
<html>
<head>
<style>
.root_td
{
background-color: #000000;
color: #FFCF00;
font-family: Verdana;
font-size: 8pt;
font-weight: bold;
height: 22;
padding-left: 5;
}
.child_td
{
background-color: #D1D1D1;
color: #000000;
font-family: Verdana;
font-size: 8pt;
font-weight: bold;
text-decoration: underline;
height: 22;
padding-left: 10;
padding-right: 10;
padding-bottom: 3;
}
body
{
color: #000000;
font-family: Verdana;
font-size: 8pt;
font-weight: normal;
}
a
{
color: #000000;
}
</style>
<script language="JavaScript">
function ShowLink(linkObject, imgObject)
{
if(linkObject.style.display == '' || linkObject.style.display == 'inline')
{
linkObject.style.display = 'none';
imgObject.src = 'plus.gif';
}
else
{
linkObject.style.display = 'inline';
imgObject.src = 'minus.gif';
}
}
</script>
</head>
<body bgcolor="#FFFFFF">
<table width="250" border="0" cellspacing="0" cellpadding="0">
<?php
while($link = mysql_fetch_array($linkResult))
{
?>
<tr>
<td class="root_td">
<img id="img_root_<?php echo $counter; ?>" onClick="ShowLink(td_root_<?php echo $counter; ?>, img_root_<?php echo $counter; ?>)" border="0" src="minus.gif" style="cursor:hand">
<?php echo $node[1]; ?>
</td>
</tr>
<tr>
<td id="td_root_<?php echo $counter++; ?>" class="child_td">
<table width="100%">
<?php
$sql = "select * from links where parentId = {$link[0]} order by title asc";
@$childResult = mysql_query($sql);
while($child = mysql_fetch_row($childResult))
{
?>
<tr>
<td class="child_td">
<a href="<?php echo $child[2]; ?>">
<?php echo $child[1]; ?>
</a>
</td>
</tr>
<?php
}
?>
</table>
</td>
</tr>
<?php
}
?>
</table>
</body>
</html>Any suggestions? Thanks