1

If I have a node object and would like to render just a given CCK field, how would I do so? I don't mean on a node template page or in a nodeapi hook, just in general.

My custom CCK field's theme_mymodule_formatter_default function takes an $elements array - when this function is called on the node page, $elements is an array with keys including the following:

  • #formatter
  • #node
  • #field_name
  • #theme
  • 0
  • 1

...where 0 and 1 point to arrays of values in the multi-valued CCK field.

Now if I want to render this field, and all I have is my formatter function and a node object, how do I format it? If I call theme('mymodule_formatter', $node->field_my_cck_field[0]), what gets passed to the $elements array of my formatter function is an array with only the following keys

  • 0
  • 1

But rather than even having to call the theme() method, I'd like to call something like drupal_render() that will pick the default theme formatter function. I imagine there's a way to do something like this, I just don't know how.

2

In D6, you can render CCK fields with:

content_format

e.g:

print content_format('field_my_cck_field', $node->field_my_cck_field[0], 'default', $node);

See also content_view_field for an alternative option.

If you're in a node template and the field is set to display, you can also just go:

print $FIELD_NAME_rendered // where $FIELD_NAME is the field name
2
$field = content_fields('field_my_cck_field', $node->type);
$rendered_field = content_view_field($field, $node, FALSE, TRUE);
  • Hello. It's usually a good idea to post a bit more than a code in your answer. A bit of explanation, links to api docs and so on. – Mołot Sep 10 '13 at 12:02
  • This is what i was looking for. Rendering a cck field with the div wrappers that they normally come with. – Bibokid Jul 30 '14 at 8:28
1

Rendering CCK Field values in Drupal 6.x

If you have a copy of the node and you need to output the rendered value of a field or fields, then you need to

  1. Set the view mode to either 'full' or 'teaser'
  2. Push the node through node_build_content
  3. Output the 'view' key of the fields you want to render

Here's an example:

<?php
$nid = 12345;
$node = node_load($nid);

if ($teaser_view_is_what_you_want) {
    $build_mode = NODE_BUILD_PREVIEW;
    $teaser = TRUE;
    $page = FALSE;
else {
    $build_mode = NODE_BUILD_NORMAL;
    $teaser = FALSE;
    $page = TRUE;
}

$node->build_mode = $build_mode;
$node = node_build_content($node, $teaser, $page);

content_alter($node);

echo $node->field_myfield[0]['view'];
?>
0

The funcitons field_view_value() and field_view_field() may help you in this case?

Have a link on this article about these functions: http://www.computerminds.co.uk/articles/rendering-drupal-7-fields-right-way

  • I should have mentioned it's Drupal 6 I'm working with. I have updated the thread title to reflect this. – user779159 Nov 11 '11 at 14:16
  • The question says D6. – doublejosh Mar 6 '12 at 7:19
  • And he says that he changed the title to d6 after i posted the answer. – Lance Mar 6 '12 at 7:50
  • whopsie daisy :) – doublejosh Apr 28 '13 at 23:54

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