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I'm having a view and I've created a template file for the row style output. I'm able to access the fields using the below code:

$fields: An array of $field objects. Each one contains:
$field->content: The output of the field.
$field->raw: The raw data for the field, if it exists. This is NOT output safe.

I'm getting output for the below code:

<?php print $fields['field_some_field_name']->content; ?>

But when I'm trying to print the fields nothing is getting returned:

<?php print_r($fields); ?>

I'm also unable to check the condition for the fields:

<?php if($fields['field_some_field_name']->content == 'Test') { } ?>
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  • Enable this module drupal.org/project/devel and then instead of print_r use dpm(); and don't use die or exit functions. Aug 12, 2016 at 11:56
  • Is the template loaded or not? Did you clear cache after adding the template?
    – Jimmy Ko
    Aug 12, 2016 at 11:57
  • @JimmyKo Yes I cleared cache after doing that. Aug 12, 2016 at 12:51

1 Answer 1

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I'm also unable to check the condition for the fields:

<?php if($fields['field_some_field_name']->content == 'Test') { } ?>

That is probably because in actuality:

$fields['field_some_field_name']->content = <div>Test</div>

Therefore, <div>Test</div> is not equal to Test

So what you could do is use raw instead, as raw has no html/markup around the value.

<?php if($fields['field_some_field_name']->raw == 'Test') { } ?>

else, you could still use content with the PHP striptags() function

<?php if(striptags($fields['field_some_field_name']->content) == 'Test') { } ?>

Or else, you could still use your original <?php print $fields['field_some_field_name']->content; ?> but as long as you strip all the markup added by views as shown in my answer to "Is there a more performant alternative than overriding template files?".

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