Try using '$id' :
SELECT properties.area, images.image_filename FROM properties,
images WHERE properties.reference_number=images.reference_number AND
properties.area='$id'
On Mon, 2003-12-08 at 23:18, fatblokeonbike wrote:
> .
> I'm new to MySQL joins with PHP and I can't get this to work - and I don't
> understand *why* it won't work.
>
> Table 1 - Countries - has fields: country
> Table 2 - Properties - has fields: area, reference_number
> Table 3 - Images - has fields: image_filename, reference_number
>
> The first page goes -
>
> $Query="SELECT country FROM countries";
> $Result=mysql_db_query ($DBName, $Query, $Link);
> while ($Row=mysql_fetch_array($Result))
> {
> print("<a href=\"country.php?id=$Row[country]\">$Row[country]</a><br>");
> }
>
> The visitor, wanting to see an area, clicks on a particular country's link
> and arrives in the country.php page, which contains -
>
> $id = $HTTP_GET_VARS["country"];.
>
> $Query="SELECT properties.area, images.image_filename FROM properties,
> images WHERE properties.reference_number=images.reference_number ";
>
> This works, but it delivers the images of every country. To call the images
> from just one country, I try -
>
> $Query="SELECT properties.area, images.image_filename FROM properties,
> images WHERE properties.reference_number=images.reference_number AND
> properties.area=$id";
>
> but it doesn't work - I get the usual "...not a valid MySQL result resource"
>
> I've played around with it, but I confess myself beat. I expect the
> answer's terribly simple - but then, as everyone keeps telling me, so am I.
>
> If you can help, thanks in advance.
>
> Iain.
--
Diana Soares
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