Can't get the URL from an image field even with the data ["url"]

• tlemaitre27

• May 17, 2017 at 12:18 pm

Hello!
I use an image field named “image_artiste”. I used image object in the option. I wanted to use the url of my image that I uploaded before to put the url as a background-image to my li element with this code: <li style="background: url('<?php the_field("image_artiste")["url"]; ?>'); "></li>
so the_field("image_artiste")["url"] doesn’t work because it gave me the complete array with all the data and not only the url even if a do an echo outsite of my li element. So i’m obligate to use the url option in the return value

thank!

• John Huebner

• May 24, 2017 at 9:52 am

the_field("image_artiste") echos the value of the field '['url'] has no meaning here since the value is not returned. Try echo get_field("image_artiste")["url"]; but I’m not sure this will work or the following which will always work.

$image = get_field("image_artiste");
echo $image['url'];