Outputting image title

• crealp

• April 10, 2014 at 12:47 am

Hi,

First of all, I would like to thank you for this great plugin, it’s really helpfull for WordPress development.

I have a problem that I can’t solve :
I have a subfiled named ‘photos’ in which I can add one or several images uploaded from WordPress.
I reach to display the image recovering the url (see the code below) but I can’t find a way to display the ‘title’ or ‘alt’ or ‘description’ of the image!!
Can you help me find the solution please ?

Here is my code :

<?php if(get_field('diaporama')){ ?>
<?php while(the_repeater_field('diaporama')): ?>
<img src="<?php the_sub_field('photos'); ?>"/>
<?php endwhile; ?>
<?php } ?>

Thanks in advance for your time.

P.S : sorry for my bad english, I’m from france 🙂

• rorymheaney

• April 11, 2014 at 2:40 pm

This will do it:

<?php  
      while(has_sub_field('diaporama')):
      $image = get_sub_field('photos');
      $image_full = $image['sizes']['large'];
      $alt = $image['title']; ?>

      <div class="col-split">
       <a class="fancybox" rel="gallery1" href="<?php echo $image_full; ?>">
       <img class="img_cen" src="<?php echo $image_thumb; ?>" alt="<?php echo $alt; ?>" title="<?php echo $alt; ?>" />
        </a>
       </div>

       <?php endwhile; ?>
   <?php endif; ?>

• crealp

• April 12, 2014 at 3:29 am

Hi rorymheaney,

Thanks for your help, this is almost what I need.

What you gave me works perfectly, but I can’t display full size image instead of large size image, even if I write that :
$image_full = $image[‘sizes’][‘full’];

Do you have an idea for that ?

Thanks again !

• crealp

• April 15, 2014 at 7:04 pm

Thanks rorymheaney,

I just replaced the line :

$image_full = $image['sizes']['large'];

By

$image_full = $image['url'];

And now everything is ok.

Thanks for your help!