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Posted by Jerry Stuckle on 04/28/06 14:21
ste wrote:
> Hi there,
>
> I'm just beginning to learn PHP and MySQL, but I'm finding it difficult! I
> wondered if someone could help me out with a problem I'm having, or at least
> point me in the right direction? I have setup a MySQL database which
> contains the following, though I would like to expand on this in the future
> with lots of extra fields:
>
> IMAGES TABLE:
> -------------------
> imageid (primary key)
> imagelocation (url location for image)
> imagecaption (caption for image, not used in the code below, but will be
> used once I suss this out!)
>
> What I want to do is, from a HTML gallery of thumbnails, be able to open a
> larger version of each thumbnail image in a nice pretty formatted HTML page.
> Each HTML page would be identical, so that's why I only want to create this
> once, as opposed to hard coding them all. As I want to keep this simple and
> one step at a time, I'm prepared to create the image gallery and the
> appropriate image URL's.
>
> My problem is that I'm unsure of the php/mysql I need to write in order to
> open the appropriate picture in the image template. For example, if I hover
> over and click 'image 1,' I would like the browser to open the page
> www.mywebsite.com/imagetemplate.php?id=1 This would open the image template
> page with image 1 visible within it.
>
> So when executing the sql query which selects a particular image from the
> database, I would like it to find the record which has an ID equal to the ID
> in the URL above (in this case, 1).
>
> Now what I've tried isn't working as I've got it all wrong, but here it is
> for interest:
>
> DODGY CODE:
> ------------------
>
> <?php
>
> /* this is the include file for my database passwords */
> include("my_db_login.inc");
>
> /* this is the code to make the connection to the database */
> $connection = mysql_connect($host,$user,$password) or die ("couldn't connect
> to server");
> $db = mysql_select_db($database,$connection) or die ("Couldn't select
> database");
>
> /* this query SHOULD be requesting all records from My Database where the
> ImageID is equal to the ImageID listed in the URL as mentioned above */
> $query = "SELECT * FROM my_database WHERE imageid =
> \"{$_POST['imageid']}\"";
> $result = mysql_query($query) or die ("Couldn't execute query.");
>
>
> /* the line of code below is actually the code from the table cell which
> SHOULD insert the image location url for the image where the id is equal to
> the one requested in the url above */
> echo "<img src=\"{$row['imagelocation']}\" width=\"500\" border=\"0\" />";
>
> ?>
>
> I'd be grateful for any help with this as I'm obviously doing something (or
> many things!) wrong.
>
> Thanks,
>
> Ste
>
>
You're close. mysql_query() returns a result object; you need to retrieve the
actual data (into $row in your example).
After the mysql_query() add the following:
$row = mysql_fetch_array($result);
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================
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