在drupal视图中用字段链接包装输出

Question

I have a view that is outputting a list of nodes that contain a link, image and text.

I want to use the link field to wrap the output (rather than link to the node), but I can't figure out how to get the raw url/title etc to create create that link in the template.

The field is configured to output the url as plain text, but is wrapped in div/spans regardless of the style settings for the view field.

views-view-fields--my-view.php:

<a href="<?php echo $fields['field_link']->content ?>">
<?php foreach ($fields as $id => $field):
    if ($id == 'field_link') continue;
    ?>
  <?php if (!empty($field->separator)): ?>
    <?php print $field->separator; ?>
  <?php endif; ?>

  <?php print $field->wrapper_prefix; ?>
    <?php print $field->label_html; ?>
    <?php print $field->content; ?>
  <?php print $field->wrapper_suffix; ?>
<?php endforeach; ?>
</a>

This produces:

<a href="<div class=" data-thmr="thmr_72"><div class="field-items"><div class="field-item even"><span data-thmr='thmr_24' class='devel-themer-wrapper'>/drupal/%237digital-buy</span></div></div></div>">
   [...]
</a>

Which is obviously not what I need.

Answer 1

Found the solution. You need to rewrite the output. See below.