I have done it this way...
$menu = array();
$count = $db->num_rows();
for($i = 0; $db->next_record(); $i++)
{
$menu[$i]["name"] = $db->f("name");
$menu[$i]["url"] = $db->f("topic_id");
}
OK, so is there a better way of acheiving the same end result?
and this is what I've got to print the results
while ( list($name, $subarray) = each($menu) )
echo "<a
href=\"".$subarray["url"]."\">".$subarray["name"]."<a><br>\n";
}
is there any better way of doing this??
thanks
M@
> -----Original Message-----
> From: Pavel Kalian [mailto:[EMAIL PROTECTED]]
> Sent: 21 February 2001 15:00
> To: Matt Williams; [EMAIL PROTECTED]; PHP_UK@egroups. com
> Subject: Re: [PHP] array headaches
>
>
> 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
>
> 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@
> >
> >
> > --
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> >
>
>
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