Trond,
What is your table structure?
I'm guessing you need a different query.
Also, be aware that mysql_fetch_array returns an array with the field name
as value, so:
$sql="select fields from table where id = $id";
$connection_handle=(mysql_connect($user, $pass, $host);
mysql_select_db($database, $connection_handle);
$result=(mysql_query($sql, $connection_handle);
$i = mysql_fetch_array($result);
echo $i[fieldname];
will produce a single result item.
Best regards,
Andrew
--------------------------------------
Andrew Hill - OpenLink Software
Director Technology Evangelism
eBusiness Infrastructure Technology
http://www.openlinksw.com
> -----Original Message-----
> From: Trond Erling Hundal [mailto:[EMAIL PROTECTED]]
> Sent: Monday, March 05, 2001 7:56 AM
> To: PHP-DB-LIST
> Subject: [PHP-DB] mysql_fetch_array problem...!
>
>
> I want to run a query to my db, fetching different fields from three
> different tables.
> In order to recognise the individual fields I give them names:
>
> select portal.portal as portal, portal.portalid as id... etc etc
>
>
> How can I refer to one specific row in this query..?
> What I mean is, how can i refer to result row number 4...?
>
> If I only selected rows from one table I could do something like this:
>
> $i = mysql_fetch_array($sql) ;
>
> echo "$i[4]" ;
>
>
>
>
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