Show content if link field is not filled

• 

  • anta

  • July 3, 2020 at 6:48 pm

  Hello,
  I have an image field and a link field.

  I want the image to link to some page within the site.
  I managed to do that. The problem comes when, because of the ‘if’ statement of the link, if the link is not filled, the image is hidden.

  How can I manage so if there’s no link, there are no pointer events and if there’s link it links to the selected page?

  Thanks,

  Here’s my code:

  	<?php if (get_sub_field('type') == "Image") { ?>
  							<div class="full-img">
  																							<?php
  																							$link = get_sub_field('link');
  																							if( $link ): ?>
  
  							<a href="<?php echo esc_url( $link ); ?>">
  										<img src="<?php the_sub_field('image'); ?>">
  										<p class="caption"><?php the_sub_field('caption'); ?>
  
  							</a>
  																						<?php endif; ?>
  							</div>
  						<?php } ?>
• 

  

  • John Huebner

  • July 3, 2020 at 10:00 pm

  
  <?php 
    if (get_sub_field('type') == "Image") {
      ?>
        <div class="full-img">
          <?php
            $link = get_sub_field('link');
            $content = '<img src="'.get_sub_field('image').'">';
            if ($link) {
              $content = '<a href="'.esc_url($link).'">'.$content.'</a>';
            }
            echo $content;
          ?>
          <p class="caption"><?php the_sub_field('caption'); ?></p>
        </div>
      <?php 
    };
  ?>
  
• 

  • anta

  • July 3, 2020 at 11:01 pm

  Thanks a lot! It worked