"if" Image Exist In A Folder.. Show It.. ?

[zchlomo]

Hello there.

 

We have just added a folder and a function to the popup_image.php file, which shows a second image in the popup window for detailed images.

 

Here are the codes:

 

<?php echo tep_image(DIR_WS_IMAGES . $products['products_image'], $products['products_name']); ?>

----- created by us to show the second image: ------

<br><br>

<?php echo tep_image('http://www.beltshop.no/catalog/images/image2/' . $products['products_image'], $products['products_name']); ?>

----------------------------------

 

What we need is a "if" rule, that tells the popup window that if there not is stored a image in the "second image folder", it should not attempt to show it.. something like this:

 

<?php echo tep_image(DIR_WS_IMAGES . $products['products_image'], $products['products_name']); ?>

 

<br><br>

"if there is a image stored in "image2" folder, show:"

<?php echo tep_image('http://www.beltshop.no/catalog/images/image2/' . $products['products_image'], $products['products_name']); ?>

 

Help anyone..

[Guest]

Hello there.

 

We have just added a folder and a function to the popup_image.php file, which shows a second image in the popup window for detailed images.

 

Here are the codes:

 

<?php echo tep_image(DIR_WS_IMAGES . $products['products_image'], $products['products_name']); ?>

----- created by us to show the second image: ------

<br><br>

<?php echo tep_image('http://www.beltshop.no/catalog/images/image2/' . $products['products_image'], $products['products_name']); ?>

----------------------------------

 

What we need is a "if" rule, that tells the popup window that if there not is stored a image in the "second image folder", it should not attempt to show it.. something like this:

 

<?php echo tep_image(DIR_WS_IMAGES . $products['products_image'], $products['products_name']); ?>

 

<br><br>

"if there is a image stored in "image2" folder, show:"

<?php echo tep_image('http://www.beltshop.no/catalog/images/image2/' . $products['products_image'], $products['products_name']); ?>

 

Help anyone..

 

You might think about using the database to your advantage if you want to do this. You can add another field in the products table that says products_image_2 and holds the location of where the second image is. That way you can do a check like

 

if($product['products_image_2'] == NULL){

--load second image here--

}

 

That's one option anyway. Otherwise you're going to have to test where the second image is and have a million other lines of code and headache to try and do the same thing.

[zchlomo]

Hmm.. is there no way around using the database? I'm not good with that at all..

 

What if I make another blank picture wich I tell the script to use if there is not stored a second image with the same name as the first image in the image2 map?