[SOLVED]rendering fields in "page--node--content_type_name.tpl.php

By adrianqx on 19 Jun 2014 at 08:54 UTC

I have a page node template named page--node--content_type_name.tpl.php
when I try calling fields using eg print render($content['field_article_author']); nothing appears , how else can I call this fields ?

Comments

For this, hide that field

preksha commented 19 June 2014 at 10:45

For this, hide that field from content like this:

  hide($content['field_article_author']);
  print render($content);

And use the render() function and the field name to print the field somewhere else:

 print render($content['field_article_author']);

If it is still not showing anything, try to debug $content array. if you have Devel module enabled you can simply place dsm($node); in the template to print out the node object.

Also check, Is the field enabled on the manage display tab for the content type?

Thanks,
Preksha

Alternative is $node variable

er.pushpinderrana commented 23 June 2014 at 11:54

Alternative is $node variable that you can use instead of $content but in special scenario. In your case first check your Manage Display setting of this content type to make sure this field is not hide from there.

Pushpinder Rana #pushpinderdrupal
Acquia Certified Drupal Expert

Hello

adrianqx commented 24 June 2014 at 07:51

Hello

<?php
 print render($content['field_article_author']); 
?>

just dosen't work it works well in node--contenttype_name.tpl.php but will not display in page--node--contenttype_name.tpl.php which I prefer using since I can alter the whole template layout as i please

Use correct variable if you

er.pushpinderrana commented 24 June 2014 at 08:03

Use correct variable if you are using page template. See Here

you can use here $node object.

Pushpinder Rana #pushpinderdrupal
Acquia Certified Drupal Expert

finally figured it out

adrianqx commented 24 June 2014 at 13:36

finally figured it out

use

<?php
 print render(field_view_field('node', $node, 'field_your_field', array('label'=>'hidden'))); 
?>

Great!

er.pushpinderrana commented 24 June 2014 at 14:19

Great!

Please edit your original thread and append [SOLVED] either in start or end of title. It would help other readers to understand the status of this post.

Thanks!

Pushpinder Rana #pushpinderdrupal
Acquia Certified Drupal Expert

finally figured it out

Cwiggo commented 4 February 2016 at 14:14

Brilliant. Worked for me too.

How to hide the empty fields with this code?

vmevada102 commented 7 July 2016 at 12:46

This code worked successfully using this above said code.

Kindly help me to hide the empty field for above said code.

Only variables should be passed by reference

jmcerda commented 5 September 2016 at 02:13

You should put that in a variable first -

    $my_value = field_view_field('node', $node, 'my_field', array('label'=>'hidden'));
    print render($my_value);

Good Job

Jabastin Arul commented 6 March 2017 at 07:21

@adrianqx: Wowwww!!!Its really worked fine for me..

Dosen't work in Drupal 7.55

seenafallah commented 21 June 2017 at 09:31

It dosen't work in Drupal 7.55
It gets me this error:

Notice: Undefined variable: node in include()
EntityMalformedException: Missing bundle property on entity of type node. in entity_extract_ids()

Any other solutions?

Almost great

Tonton2 commented 4 July 2018 at 13:11

The last solution kind of worked for me. I am using Drupal 7.59
The image shows up but I get an error message: Strict warning : Only variables should be passed by reference

Right, variable only should be passed.

Tonton2 commented 5 July 2018 at 14:38

I watched how I did on another site earlier.

<?php if (!empty($node)):
	$fieldImage = field_view_field('node', $node, 'field_image', array('label'=>'hidden'));
	if ($fieldImage): ?>
		<?php print render($fieldImage);
	endif;
endif; ?>