How can I echo the url of the image from the selected field?

Kirby 2 Solved ✅
1

Hi,
I need a structured field for employees!
Photo, employee name, job title and phone.

It works so far - I can even select between different images - but how can I echo the url of the image from the selected field in my template?

Thanks for help!
Dan

My template

<?php $addresses = yaml($page->contact()) ?>
<?php foreach($addresses as $address): ?>
    <div class="employee-photo">
        /* place image here - this does not work: */
        <?php if($photo = $page->image($page->photo())) :?>
            <img src="<?php echo $photo->url() ?>" />
        <?php endif ?>
    </div>
                    
    <div class="address">
          <?php echo $address['name'] ?><br />
          <?php echo $address['job'] ?><br />
        <?php echo $address['phone'] ?>
    </div>
<?php endforeach ?>

My blueprint:

title: Pure
pages: true
files: true
fields:
  title:
    label: Titel
    type:  text
  subtitle:
    label: Subtitel
    type:  text
  text:
    label: Text
    type:  textarea
  contact:
    label: Contact
    type: structure
    entry: >
      {{photo}}<br />
      {{name}}<br />
      {{job}}<br />
      {{phone}}
    fields:
      photo:
        label: Photo
        type: select
        options: images
      name:
        label: Name
        type: text
      job:
        label: Job
        type: text
      phone:
        label: Phone
        type: text
2

This should work:

$page->image($address['photo'])->url();
1 Like
3

Ah ok - thank you!!!

So in my case:

<div class="employee-photo">
     <img src="<?php echo $page->image($address['photo'])->url(); ?>" alt="" />
</div>