Re: [PHP-DB] data from db to a page and then to another page

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On Wed, Jan 7, 2009 at 4:54 AM, Mika Jaaksi <mika.jaaksi@xxxxxxxxx> wrote:

> I already can get the data from database to a page. Now I want to make link
> from that data to next page and on that new page it should show all the
> data
> that is related.
>
> example:
>
> data from database
> -->
> page1 where listed:
>
> band1 (a href)
> band2 (a href)
> band3 (a href)
> ...
>
> and when clicking for example band3
> -->
> page2 where listed band info:
>
> bandname
> bandhistory
> bandmembers
> ...
>
> So, how should I do this? Should I somehow use $_POST method to
> send/deliver
> band_id to the next page?
>

You could do several things.

1) Use javascript to make a link using the band3 value within the Javascript
to pass the ID to the next page.

Example:

Have a Javascript function -

<script language="JavaScript">
<!--
function openWin(band_id) {
 var link
 link = my/directory/display.php?band_id=' + band_id;
 MyWin = window.open(link,"OpenPage");
 MyWin.focus();
}
//-->
</script>

And then call that function when you click on the band you want.

<a href="javascript:openWin('<?php echo $band_id; ?>')"><?php echo $band;
?></a>

2) You could use form objects (radio buttons/check boxes/dropdown box etc)
to pass the band_id value via POST

Then on your second page all you would have to do is get the value.

Example:
<input type="radio" tabindex="1" name="band" value="<?php echo $band_id;
?>">
<input type="radio" tabindex="2" name="band" value="<?php echo $band_id;
?>">
<input type="radio" tabindex="3" name="band" value="<?php echo $band_id;
?>">

On the second page you would simply check the posted value of you "band"
radio input:

$band_id = $_POST['band'];

Hope that helps.

Dan

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