Display a block on some pages according to the fields content of that page

Question

I created a content type called people, where I create personal pages using that type. One of the fields is 'area'.

Then, I created a view of block called 'software people', where I list all the people who work in software area. I want to show this block only on the pages where the person's area is Software. How can I achieve this?

Thanks

Answer 1

This goal can be achieved in views as well, but only with some PHP contextual filter code:

Just add contextual filter: Content: area (field_area)

Under: "When the filter value is NOT available" select:

Provide default value-> PHP Code

Enter PHP code:

$node = menu_get_object();
return isset($node->field_area['und'][0]['value']) ? 
  ($node->field_area['und'][0]['value']) :'';

Under: "When the filter value IS available or a default is provided" select:

Specify validation criteria->PHP Code

$area = $handler->argument;   // content of field area
return ($area == 'Software'); // validation of the area content

action to take if filter value does not validate -> hide view

That's it! Your block will show up only on the pages where the person's area is 'Software'

Answer 2

Simplest way is to conditionally output region in node.tpl.php But there is no $page in node.tpl.php. So you need to make it happen. In your template.php add function

mytheme_preprocess_node(&$variables) { 
  // List of all regions for the current theme
  foreach (system_region_list($GLOBALS['theme']) as $region_key => $region_name) {
    // Add content to $region variable
    if ($blocks = block_get_blocks_by_region($region_key)) {
      $variables['region'][$region_key] = $blocks;
    }
    else {
      $variables['region'][$region_key] = array();
    }
  }
}

Then in node.tpl.php use:

print render($region['region_name']);

to output your region. Of course wrap it in if, case or any other logic you fancy.