> a) $menu[] = array("name" => $db->f("name"), "url" => $db->f("topic_id"));
>
> b) echo $menu[0]["name"]; //for example - depends on what you want to do
Thanks, maybe I should have made myself a little clearer though
I am expecting more than one row to be returned from the db.
So how would I get the all the rows returned into an associative array??
The manual tells you how how it should, but I need to do it from
mysql_fetch_array
M@
>
> BTW have a look into the manual, the array stuff is described pretty well
>
> Pavel
>
> ----- Original Message -----
> From: "Matt Williams" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>; "PHP_UK@egroups. com"
> <[EMAIL PROTECTED]>
> Sent: Wednesday, February 21, 2001 3:56 PM
> Subject: [PHP] array headaches
>
>
> > Hi
> >
> > could anybody please help me get my head around the following.
> >
> > I get some url information from the database (title, url)
> > I am trying to then put this into an array so I can pass it to
> the page to
> > display a menu.
> > so
> > a) how do I get the data into the array. would this be the way
> to do it??
> >
> > $menu = array();
> > while($db->next_record())
> > {
> > $menu[]["name"] = $db->f("name");
> > $menu[]["url"] = $db->f("topic_id");
> > }
> > which leads me onto
> > b) how do I get the data back out. I can't test to see if the
> above works
> as
> > I can't get at the data
> >
> > TIA
> >
> > M@
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
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> > To contact the list administrators, e-mail: [EMAIL PROTECTED]
> >
>
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