Problems Creating the Code to Open Images Within a Template PHP Page

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

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