[RESOLVED] Image Gallery from dynamic folder with Thumbnails

[RobertSmith]

Hello,
I want to have an image gallery from a dynamically created folder. I have a working script that takes the images from a static location. But somehow unsuccessful in getting the database folder name for dynamic linking.:mad:
The code I am on is as below

<?php
$a = '0';

$filepath = "thumbs";
$url_path = "images";
$dir = dir($filepath);
echo "<table border=\"0\" cellpadding=\"5\" cellspacing=\"5\" width=\"75%\">";
while($entry=$dir->read()) {
if($entry == "." || $entry == "..") {
continue;
}
$fp = @fopen("$filepath/$entry","r");
if ($a == '0') {echo "<tr>";}
if ($a == '5') {echo "<tr>";}
if ($a == '10') {echo "<tr>";}
if ($a == '15') {echo "<tr>";}
?><td>
<a href="<?php echo "$url_path/$entry" ?>">
<img src="<?php echo "$filepath/$entry" ?>" alt="<?php echo $entry ?>"></a>
</td>

<?php
$a = $a + 1;
}
?>
</tr>
</table>
</center></div>
</body>
</html>

While with database query the image path to be something like "<?php echo $row_Recordset1'[IMG_ID]'; ?>"

and Thumbnail path "<?php echo $row_Recordset1'[T_ID]'; ?>"

But with this query I am getting error
Fatal error: Call to undefined function <?php echo Array'[T_ID]'; ?>() in C:\wamp\www\slideshow\gallery9.php on line 56

Where I am going wrong??
Thanks

[nrg_alpha]

<?php echo Array'[T_ID]'; ?>() looks like an issue..

a) Your array should be Array['T_ID'] (the quotes on the inside of the square brackets..)
b) You have () afterwards.. this is not a proper function.. I'm not sure what you are trying to do with the () part... As it stands, it is useless.. unless you are trying to call a function (in which case, this syntax is not correct), unless you have something like:

function <?php echo Array['T_ID']; ?>(), in which case this only works if the value of Array['T_ID'] is a function name that exists..

EDIT - Please note that the code you cut and paste into these forums should be surrounded by [ php] and [ /php] tags (without any spaces in them). This will present your code with colour coded key syntax words (as found in IDEs) which makes things more easily readable.

[RobertSmith]

Thanks nrg_alpha for your quick reply.
I have corrected the quotes part.
I am getting the <?php echo Array['T_ID']; ?>() as part of the error message on the browser.

function <?php echo Array['T_ID']; ?>(), in which case this only works if the value of Array['T_ID'] is a function name that exists..

Sorry, I don't have any clue on this. What I am trying to echo (is that the right term) two database entries in the place of the image and thumbnail path. respectively. I am able to echo the database entries as a text in the page, which means the query is working fine. But the probelm is getting it in the $filepath and $url_path.

Being a first timer I didn't know how to put the code. Thanks for your help on this. I am posting the code as you said.

<?php
$a = '0';

$filepath = "thumbs";
$url_path = "images";
$dir = dir($filepath);
echo "<table border=\"0\" cellpadding=\"5\" cellspacing=\"5\" width=\"75%\">";
while($entry=$dir->read()) {
if($entry == "." || $entry == "..") { 
continue; 
} 
$fp = @fopen("$filepath/$entry","r");
if ($a == '0') {echo "<tr>";}
if ($a == '5') {echo "<tr>";}
if ($a == '10') {echo "<tr>";}
if ($a == '15') {echo "<tr>";}
?><td>
<a href="<?php echo "$url_path/$entry" ?>">
<img src="<?php echo "$filepath/$entry" ?>" alt="<?php echo $entry ?>"></a>
</td>


<?php
$a = $a + 1;
}
?> 
</tr>
</table>
</center></div>
</body>
</html>

[nrg_alpha]

RobertSmith;10889014 wrote:

I am getting the <?php echo Array['T_ID']; ?>() as part of the error message on the browser.

You need to remove those trailing parenthesis.

[igorlopez]

I tested your code as posted and it worked as expected(I am getting the images in rows with 5 in each row with the url as <webroot>/images/<filename>) but I had to change the filepath to

$filepath = "./thumbs";

for it to work but I am running Linux so it got probably nothing to do with your problem.
I removed the error suppression operator before fopen but it worked both with and without.

[RobertSmith]

nrg_alpha, this is where I am confused.

You need to remove those trailing parenthesis.

I do not have this parenthesis in the code. Where from I will remove this? I getting this error and parenthesis while I test in the browser. Sorry for being dump.

igorlopez, the code as such is working fine. But the problem here is while targeting the images and thumbnail folder to a database field. I want to replace this part

$filepath = "thumbs"; 
$url_path = "images"; 

with something like this

$filepath = <?php echo $row_image_query'[T_ID]'; ?>

and

$url_path = <?php echo $row_image_query'[IMG_ID]'; ?>

where, T_ID and IMG_ID will point to a dynamic folder

Thanks

[nrg_alpha]

RobertSmith;10889105 wrote:

nrg_alpha, this is where I am confused.

I do not have this parenthesis in the code. Where from I will remove this? I getting this error and parenthesis while I test in the browser.

What does the error and code on line 56 now look like (assuming the error still reflects this line..otherwise, show the line of code that the error is referring to)?

I'm lost.

Internet communication can be so confusing at times... :queasy:

[igorlopez]

My turn to ask,
What do you want to achieve with this:

$filepath = <?php echo $row_image_query'[T_ID]'; ?>
$url_path = <?php echo $row_image_query'[IMG_ID]'; ?>

It would be interesting to just see the following echo:

// Replace this:
//$filepath = "thumbs";
//$url_path = "images";
// with this and show us the resulting echos
echo $row_image_query'[T_ID]';
echo $row_image_query'[IMG_ID]';

[nrg_alpha]

Robert, using this line as an example:

$filepath = <?php echo $row_image_query'[T_ID]'; ?>

You have to move the quotes to the inside of the square brackets:

$filepath = <?php echo $row_image_query['T_ID']; ?>

I would comb through your code and fix those.

[RobertSmith]

Hi all.
Thanks for your help so far. I am sort of stuck with this for the past 2 weeks or so. I think I will have to re-frame the question now. Let me explain a bit. My client needs to display various images of same product in the product detail page. The problem here is that the number of images for a single product can vary. Some times it can be just 4 and at time it can be up to 14. So I decided to set up a multiple file upload feature (using ADDT) and save the folder path to MySQL, something like 'Image_Folder_Path'. I tried innumerable gallery scripts, but unfortunately none of them support dynamic folder paths.

Below is such a code I found at http://www.ricocheting.com/scripts/gallery.html
{I wanted to copy the entire script here, but got message that I can't go more than 10000 chars. I am stuck again!!)

If you get a chance to look at the script......

What I want is to change is

$config['fulls'] = "photos/";
$config['thumbs'] = "thumbs/";

to a dynamic database entry.

I was trying to call this as below

$config['fulls'] = "<?php echo $row_Recordset1['Main_Img']; ?>/";

But with no avail, where I am going wrong.
The database results echoes successfully when called in the body.
Thanks for any help

[igorlopez]

I am most likely very thickheaded but can you explain to me why you want to put

"<?php echo $row_Recordset1['Main_Img']; ?>/"

into a php variable?
As I have no idea where you intend to use this $config['fulls'] variable more then from your example code,

<a href="<?php echo "$url_path/$entry" ?>">
<img src="<?php echo "$filepath/$entry" ?>" alt="<?php echo $entry ?>"></a>

Why not use it in the same way but exchange $filepath/$entry with $row_Recordset1['Main_Img'] depending of course what is inside row_Recordset1.
Again, if you can show us what your intention is and what is inside $row_Recordset1['Main_Img'] we might be of more help.

You said earlier the you wanted to replace

$filepath = "thumbs"; 

with

$filepath = <?php echo $row_image_query'[T_ID]'; ?>

but maybe your meaning is that

$filepath = $row_image_query['T_ID'];

and then use it in your display section

<img src="<?php echo "$filepath/$entry" ?>" alt="<?php echo $entry ?>"></a>

Sorry for just rambling and not have any solution.

[RobertSmith]

Hi igorlopez,

$filepath = $row_image_query['T_ID'];

does the trick. Now I can see it working. Thanks a billion. You saved me for now!!!🙂
Thanks, again.