does this look right? still learning

[computerbarry]

hi all

i have this bit of code which is selecting an img and showing on my viewphoto.php

it works, but it just dosent look right ive seen this type before but the .jpg does not show in the address bar to view the image

http://mysite.com/viewphoto.php?file=images/paddy/patrick11.jpg&photo_id=11

would it be something like

http://mysite.com/viewphoto.php?file=images&photo_id=11

????

if that makes sense
thanks!!!

[php_daemon]

If the photo_id is unique for a single image, that's all you should need for the params in url:

http://mysite.com/viewphoto.php?photo_id=11

The first method is not only redundant but insecure too. Anyone could change the path and access the files you do not intend to make available.

The second method is redundand for the same reason. Although now you remove the security risk, still there's no need for multiple identification.

Hope this helps.

[computerbarry]

yes tought so, thanks php daemon!!!

so

http://mysite.com/viewphoto.php?photo_id=11

is the safe option?

[computerbarry]

ive been trying to alter this but when i do the img cant be found(well just dosent show).

echo '<td><a href ="viewphoto.php?file=images/paddy/' . $row['image'] . '&photo_id='. $row['photo_id'].'"> 
<img src ="thumbs/paddy/' . $row['image'] . '"></a><br /><p>' . $row['caption'] . '<br /><span class="orange">Views:' . $row['views'] . '</span></p></td>'; 

the . $row['image'] . has the patrick01.jpg in it, i deleted this

and tryed this

echo '<td><a href ="viewphoto.php?photo_id='. $row['photo_id'].'"> 
<img src ="thumbs/paddy/' . $row['image'] . '"></a><br /><p>' . $row['caption'] . '<br /><span class="orange">Views:' . $row['views'] . '</span></p></td>'; 

which worked but didnt show the img

any suggesstions???

thnaks!!!!!!!

[php_daemon]

Does $row['image'] really contain patrick01.jpg now when you changed the url? Post the full script, something may be in the sql query or elsewhere.

[computerbarry]

the url now:

http://mysite.com/viewphoto.php?photo_id=8

$row['image'] is a field in my table where ive stored all the file names to refer to.

thanks

this is the viewphoto.php

$file = $_GET['file'];
$photo_id = $_GET['photo_id'];
$update = mysql_query("UPDATE photograph SET views = views+1 WHERE photo_id = '".mysql_real_escape_string($photo_id)."'") or die(mysql_error()); 
echo '<img src="'.$file.'">'; 

if that helps

[php_daemon]

I see. You used to use photo_id to update the views and file to display the image. Well, you should use photo_id for both, that'll make it two queries:

$photo_id = $_GET['photo_id']; 
$update = mysql_query("UPDATE photograph SET views = views+1 WHERE photo_id = '".mysql_real_escape_string($photo_id)."'") or die(mysql_error()); 
$resource = mysql_query("SELECT image FROM photograph WHERE photo_id = '".mysql_real_escape_string($photo_id)."'") or die(mysql_error()); 
echo '<img src="'.mysql_result($resource,0,"image").'">'; 

[computerbarry]

yes thats right, but still dosent show the img on viewphoto.php

what r the changes needed to this , i think this is the problem now.

echo '<td><a href ="viewphoto.php?file=images/paddy/' . $row['image'] . '&photo_id='. $row['photo_id'].'"> 
<img src ="thumbs/paddy/' . $row['image'] . '"></a><br /><p>' . $row['caption'] . '<br /><span class="orange">Views:' . $row['views'] . '</span></p></td>'; 

thanks php daemon

[php_daemon]

Hm, is there an image file under thumbs/paddy/? Because if the script outputs the image file name into HTML, then the only problem is the absence of the image in the said directory.

[computerbarry]

photodir.php is my main photo page which shows thumbs/paddy/ = all the img thumbs which have a link to viewphoto.php

and when clicked displays viewphoto.php?photo_id=8 with the bigger img like weve been doing. if u get what i mean?

[php_daemon]

OK. Something's wrong with the query in photodir.php then. Can you post that code, please.

[computerbarry]

cheers! here the full code:

require_once('includes/photo_mysql_connect.inc.php'); 

#make the query 
$sql = 'SELECT photo_id , image, caption, views
FROM photograph WHERE gallery = 1';
$result = @mysql_query ($sql)or die('Error: ' . mysql_error());
                 
echo '<div id="viewphoto"><h4>St. Patricks Day</h4><table><tr>'; 
         
for ($i = 0; $row = mysql_fetch_array ($result); $i++) 
{ 
// close and start a new paragraph after every 4th img 
if (intval($i / 4) == ($i / 4)) 
echo '</tr><tr>'; 

echo '<td><a href ="viewphoto.php?file=images/paddy/' . $row['image'] . '&photo_id='. $row['photo_id'].'"> 
<img src ="thumbs/paddy/' . $row['image'] . '"></a><br /><p>' . $row['caption'] . '<br /><span class="orange">Views:' . $row['views'] . '</span></p></td>';
    } 
     
echo '</tr></table></div>'; 

thanks!!!!!!!!!!!

[php_daemon]

So you're saying that it does not display the thumbnails? But it doesn't have to do anything with the url that you're playing with. Or am I missing something?

[computerbarry]

no, the thumbnails display ok along with the caption and viewed times, it's when i click the thumbnails which opens up the <a href ="viewphoto.php?file=images/paddy/

thats were the images/paddy/ comes from in the address bar along with . $row['image'] which shows patrick01.jpg and so on

[php_daemon]

Oh, I didn't get you right at first. And I missed something when I changed your code for viewphoto.php. Try this:

$photo_id = $_GET['photo_id']; 
$update = mysql_query("UPDATE photograph SET views = views+1 WHERE photo_id = '".mysql_real_escape_string($photo_id)."'") or die(mysql_error()); 
$resource = mysql_query("SELECT image FROM photograph WHERE photo_id = '".mysql_real_escape_string($photo_id)."'") or die(mysql_error()); 
echo '<img src="images/paddy/'.mysql_result($resource,0,"image").'">'; 

[computerbarry]

yes that works, but it just what i had in the first place:

http://mysite.com/viewphoto.php?file=images/paddy/patrick08.jpg&photo_id=8

dont we need to change this sum way?

echo '<td><a href ="viewphoto.php?file=images/paddy/' . $row['image'] . '&photo_id='. $row['photo_id'].'"> 
<img src ="thumbs/paddy/' . $row['image'] . '"></a><br /><p>' . $row['caption'] . '<br /><span class="orange">Views:' . $row['views'] . '</span></p></td>'; 

thanks daemon

[php_daemon]

Yes, remove the file part from the url and it's good:

echo '<td><a href ="viewphoto.php?photo_id='. $row['photo_id'].'"> 
<img src ="thumbs/paddy/' . $row['image'] . '"></a><br /><p>' . $row['caption'] . '<br /><span class="orange">Views:' . $row['views'] . '</span></p></td>'; 

[computerbarry]

thats the one!!!!!!!!! CHEERS php daemon!!!

THANKS!

got there in the end

[php_daemon]

You're welcome.