php image-gallery

[Manwe]

ok,
so I'm working on my new homepage, and I decided I'd replace the Expose flash-gallery I'd make my own gallery, to have the page completly 'homemade'...
Though my knownledge on php isn't too good, I started, and I think I've gotten quite far, now what I'm struggeling with is a category menu at the top, so that I can organize my pictures into categories, without having to make a new html document for each category...

This is how far I've gotten:

<?php

class Gallery {

var $userstable = "gallery";
var $result;

function DoGallery(){

mysql_connect('localhost','root','')
|| die(mysql_error());
@mysql_select_db('thorbear')
|| die("Unable to select database");

if(!isset($_GET['userstable'])) $notgallery = "notgallery";
	echo '<a href="?userstable='.$notgallery.'"> category </a>';

$query = "SELECT * FROM $this->userstable";
$this->result = mysql_query($query);
$number = mysql_num_rows($this->result);
$i = 0;
if ($number == 0) :
	print "<center><b>No Images? O.o</b></center>";
elseif ($number > 0) :
	while ($i < $number):
		$path = mysql_result($this->result,$i,"path");
		$thumb = mysql_result($this->result,$i,"thumb");
        $alt = mysql_result($this->result,$i,"alt");
		print	"<div class=\"img\">";
		print	"<a href=\"$path\" target=\"_new\">";
		print	"<img src=\"$thumb\" alt=\"$alt\" width=\"110\" height=\"90\" style=\"-moz-opacity:0.4;filter:alpha(opacity=40)\" onmouseover=\"this.style.MozOpacity=1;this.filters.alpha.opacity=100\" onmouseout=\"this.style.MozOpacity=0.4;this.filters.alpha.opacity=40\" />";
		print	"</a>";
		print	"<div class=\"desc\">$alt</div>";
		print	"</div>";
        $i++;
	endwhile;
endif;

mysql_free_result($this->result);
mysql_close();

}
}

$p_obj =& new Gallery;
$p_obj->DoGallery();

?>

the images show properly, just the way I want them to, but I can't make the menu work...any tips? (other improvements are accepted with thx )

thx in advance

[Availor]

Manwe, I cannot help you with the code but I can give you some links to free php galleries, perhaps you can play around and see what you need from there.

[Manwe]

Manwe, I cannot help you with the code but I can give you some links to free php galleries, perhaps you can play around and see what you need from there.

nah, i have some of those, they aren't too helpful, cuz i know too little about php, and they're usually made to fit with everything...
I'd prefer some more direct help...
but thx anyway

[stlewis]

Hi

It looks as if you have only one table - 'gallery' - when really you need two - one would be called 'gallery' and the other 'image'. So each image has a foreign key 'galleryid' to link it back to the 'gallery'. The 'gallery' table would give you the gallery names.

So you then have two chunks of php - one to deliver the names of the galleries, and a second to get the images for the selected gallery.

[Manwe]

Sounds fair
but I'm not sure if I understood you 100%

1: the two tables, 1 for each image and 1 for the gallery...but I didn't quite get the reason for it...

2: the two php pieces, that is what I've attempted, but obviously not succeded

maybe you could make a very simple example? (VERY simple, just to kick me off in the right direction )

PS. I was planning to work on this today and look closer on ur tip, but I've had as much as 10 min by the comp until now, and now it's 11pm...I'll look closer onto it tomorrow, maybe I'll understand it better then...

[stlewis]

Hi

I've attached a diagram...

The first script, gallerylist.php, gets the gallery data from the gallery table and creates the menu. Each entry in the list is an <a href=...> calling the showimage.php script and passing to it the galleryid.

The showimage.php script can now do a query to get the data for the images in that gallery.

Why use two tables? Because a) it is good practice and b) because with a single table the gallery data has to be stored for every image which means that updating it becomes unecessarily complicated.

For an example of how it looks, try http://www.theeventsbusiness.com/gallerylist.php

[stlewis]

Hi

Would this help: http://www.sitepoint.com/article/php...system-minutes ?

If your folder structure is say:

pictures
- 123.jpg
- 124.jpg
php
- gallery.php

Then gallery.php must echo with ../pictures/ and then the filename.

[Suvek]

echo "\n\t\t<a href=\"purePHP.php?\$galleryid=$GID\">$name</a><br />";

These links don't work as expected, am I correct?

"$" symbol is not used in query string variable names but you use "\$galleryid". That won't work Try changing it to this:

echo "\n\t\t<a href=\"purePHP.php?galleryid=$GID\">$name</a><br />";

And you may also need to change this MySQL query (depends on your PHP configuration):

$query2 = mysql_query("SELECT * FROM image WHERE GID = '$galleryid'");

to

$query2 = mysql_query("SELECT * FROM image WHERE GID = '".$_GET['galleryid']."'");

[remi]

Hi Suvek,

that's insecure code:

$query2 = mysql_query("SELECT * FROM image WHERE GID = '".$_GET['galleryid']."'");

Such a code is open to SQL injections, please don't do that.

see also: Wikipedia article about SQL injection

Remi

[Suvek]

code is open to SQL injections, please don't do that.
Remi

Right, I should have mentioned that Thanks
But to make it clear (well it is probably obvious ) - SQL injection is a threat in both code versions - doesn't matter if you access that GET var as $galleryid or $_GET['galleryid']. My original point that $galleryid may not work directly remains the same.