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I have a content type "Person" which has a field called "state". On the node page of any "person", I want to display other nodes of "Person" which has the same value for the "state" field.
For e.g. If a we are on a person node who is from New York, the view block placed on this node should display other persons from New York.

I know how this could be done if it were node id, or term id : using contextual filters and getting content id from URL, etc. However, I'm not sure how do I get this from other text field like "state".

Any pointers would be appreciated.

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I think you need use the field type "list" or the field type "Taxonomy term".

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    Thanks for your reply, in our project we have to taken field type "text". so can you pls suggest any option using field type "text". – Rahul Baisane Sep 16 '14 at 8:35
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I don't know if it is possible with Views UI. But you could always recover your "field_state" value manually by getting the current node (assuming you're running the code in a node page):

$node = menu_get_object();
$state = $node->field_state[LANGUAGE_NONE][0]['value'];

And then use that value to look for other similar nodes, for example with EntityFieldQuery:

$query = new EntityFieldQuery();
$query->entityCondition('entity_type', 'node')
  ->entityCondition('bundle', 'person')
  ->fieldCondition('field_state', 'value', $state, '=');
$results = $query->execute();

And display the results with theme_table maybe.

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