It means your query failed.
Change
$news = mysql_query('select * from ccl where '.$where.' order by AU desc');
TO:
$news = mysql_query('select * from ccl where '.$where.' order by AU desc')
or die("error: ".mysql_error());
this will display the error
-----Original Message-----
From: jtjohnston [mailto:[EMAIL PROTECTED]]
Sent: Monday, February 18, 2002 12:30 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Frustrating ?
Heres's a frustrating, and maybe not so stupid question?
I'm getting "Warning: Supplied argument is not a valid MySQL result
resource" on this line:
while ($mydata = mysql_fetch_object($news))
So what am I doing wrong here:
$where = "id like $id";
$news = mysql_query('select * from ccl where '.$where.' order by AU desc');
//desc => z-a
while ($mydata = mysql_fetch_object($news))
{
echo "<tr bgcolor=\"#CCCCCC\"><td align=center><a
href=\"index.html?id=$mydata->id\">Print
View</a></td><td>$mydata->id</td><td>$mydata->AU</td><td>$mydata->ST</td><td
>$mydata->BT</td></tr>\n";
}#end of while
I've tried variations like:
$news = mysql_query('select * from ccl where id like $id order by AU
desc');
$news = mysql_query("select * from ccl where id like $id order by AU
desc");
$id checks out ok. I use index.html?id=4
AU exists; ok. So ... ?
John
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