populate feed even if no images in PHP - php

Question

I need to populate an ecommerce feed with extraimages
here is the header :
|image_1|image_2|image_3|
I've a table where I store or not, extra images for one product.
So I make a request to see if there are some extra images, with an sql limit of 3 as I only want 3 images max in my feed.
What I need is to populate the feed with empty pipes if there are no images and I don't know the logic to do that.
The output would be, for example, like this if I have 2 extra images :
|url/image_1.jpg|url/image_2.jpg|| <- empty pipe to match the header of 3
or only one image :
|url/image_1.jpg|||
My code so far :
$products_extra_images_query = tep_db_query(
"SELECT products_extra_image, products_extra_images_id
FROM " . TABLE_PRODUCTS_EXTRA_IMAGES .
" WHERE products_id='" . (int)$row->id . "' LIMIT 3");
if (tep_db_num_rows($products_extra_images_query) >= 1){
// there are some extra_images
while ($extra_images = tep_db_fetch_array($products_extra_images_query)) {
$output .= $extra_images['products_extra_image']."|" ;
}
}
thanks for your help.
Sebastien

Answer 1

Create an array of 4 and put your image path into it, then implode it into a string separated by "|".
$images = array_fill(0,4,null);
$idx = 0;
if (tep_db_num_rows($products_extra_images_query) >= 1) {
// there are some extra_images
while ($extra_images = tep_db_fetch_array($products_extra_images_query)) {
$images[$idx] = $extra_images['products_extra_image'];
$idx++;
}
}
$output .= implode($images, '|');

Answer 2

If I understood correctly then you could try an approach like this ~ though using non-standard names such as tep_db_fetch_array perhaps confuse the issue?
$rows = tep_db_num_rows( $products_extra_images_query );
if ( $rows > 0 ){
$output=[];
/* add found records to the `output` array rather than a string - or, use `fetch_all` */
while( $extra_images = tep_db_fetch_array($products_extra_images_query)) {
$output[]=$extra_images['products_extra_image'];
}
/* an odd looking loop but should add extra `pipe` chars when there are not 3 results */
for( $i=0; $i < ( 3 - $rows ); $i++ )$output[]='';
echo '|' . implode('|',$output) . '|';
}