Jump to content

Displaying all images from a page on a listing


Jon E
 Share

Recommended Posts

Hi,

I want to display all of the images from a page on the listing page - so for example I want the markup to be similar to:

<?php include 'head.inc'; ?>

<div id="portfolio_list">
        
<?php

$features = $pages->find("template=project, sort=-date");

foreach($features as $feature) {

   $thumb = $feature->thumbnail ? $feature->thumbnail->width(800)->url : "http://i.imgur.com/1zxfxMW.jpg"; // if no thumbnail, replace by placeholder

   echo "<ul class='bxslider'>" .
"<li><img src='{$thumb}' /></li>
  <li><img src='{$thumb}' /></li>
  <li><img src='{$thumb}' /></li>
  <li><img src='{$thumb}' /></li>
</ul>" . //changed from $thumb->url to only $thumb
";

}

?>

</div>

<?php include 'foot.inc'; ?>

...roughly.

Could anyone help me out?

Thanks

Link to comment
Share on other sites

bluewithoutgreen, you didn't tell us what the problem is, so I just can guess. As far as I can see, your code outputs the same thumbnail four times, and you get as many unordered lists as you have got pictures in $feature->thumbnail.

I think what you really want to do is:

$output = '';  // holds the HTML to output

foreach($features as $feature) {
    $thumb = $feature->thumbnail ? $feature->thumbnail->width(800)->url : 'http://i.imgur.com/1zxfxMW.jpg'; // if no thumbnail, replace by placeholder
    $output .= '<li><img src="' . $thumb . '" alt="" /></li>';
}

echo '<ul class="bxslider">' . $output . '</ul>';
  • Like 1
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...