Drupal db_select (), how can I use two db fields in state?

Question

I have a problem with drupal db_select.

Here is my code:

$query = db_select('node', 'n'); $query->addField('n', 'nid', 'nid'); $query->addField('cfs', 'entity_id', 'feature_support_id'); $query->addField('fpffs', 'entity_id', 'parent_feature_support_id'); $query->addField('cfsfc', 'feature_support_compared_target_id', 'feature_support_compared'); $query->addField('fpffsfc', 'feature_support_compared_target_id', 'parent_feature_support_compared'); //Get feature_support of the feature $query->join('field_data_feature_support_feature', 'cfs', 'n.nid = cfs.feature_support_feature_target_id'); $query->join('field_data_feature_support_compared', 'cfsfc', 'cfs.entity_id = cfsfc.entity_id'); //Get parent feature_support through feature $query->join('field_data_feature_parent_feature', 'fp', 'n.nid = fp.entity_id'); $query->join('field_data_feature_support_feature', 'fpffs', 'fp.feature_parent_feature_target_id = fpffs.feature_support_feature_target_id'); $query->join('field_data_feature_support_compared', 'fpffsfc', 'fpffs.entity_id = fpffsfc.entity_id'); $query->condition('n.nid', $node_revision->nid, '='); $query->condition('cfsfc.feature_support_compared_target_id', 'fpffsfc.feature_support_compared_target_id', '='); $result = $query->execute(); 

In the DB, my query should be

 SELECT n.nid AS nid, cfs.entity_id AS feature_support_id, fpffs.entity_id AS parent_feature_support_id, cfsfc.feature_support_compared_target_id AS feature_support_compared, fpffsfc.feature_support_compared_target_id AS parent_feature_support_compared FROM node n INNER JOIN field_data_feature_support_feature cfs ON n.nid = cfs.feature_support_feature_target_id INNER JOIN field_data_feature_support_compared cfsfc ON cfs.entity_id = cfsfc.entity_id INNER JOIN field_data_feature_parent_feature fp ON n.nid = fp.entity_id INNER JOIN field_data_feature_support_feature fpffs ON fp.feature_parent_feature_target_id = fpffs.feature_support_feature_target_id INNER JOIN field_data_feature_support_compared fpffsfc ON fpffs.entity_id = fpffsfc.entity_id WHERE (n.nid = '9') AND (cfsfc.feature_support_compared_target_id = fpffsfc.feature_support_compared_target_id) 

This query works when I try to use it in phpmyadmin, but instead in mysql log I have

 SELECT n.nid AS nid, cfs.entity_id AS feature_support_id, fpffs.entity_id AS parent_feature_support_id, cfsfc.feature_support_compared_target_id AS feature_support_compared, fpffsfc.feature_support_compared_target_id AS parent_feature_support_compared FROM node n INNER JOIN field_data_feature_support_feature cfs ON n.nid = cfs.feature_support_feature_target_id INNER JOIN field_data_feature_support_compared cfsfc ON cfs.entity_id = cfsfc.entity_id INNER JOIN field_data_feature_parent_feature fp ON n.nid = fp.entity_id INNER JOIN field_data_feature_support_feature fpffs ON fp.feature_parent_feature_target_id = fpffs.feature_support_feature_target_id INNER JOIN field_data_feature_support_compared fpffsfc ON fpffs.entity_id = fpffsfc.entity_id WHERE (n.nid = '9') AND (cfsfc.feature_support_compared_target_id = 'fpffsfc.feature_support_compared_target_id') 

See at the end, in WHERE, there is one quote around 'fpffsfc.feature_support_compared_target_id', which should not be.

This is obvious because the second argument of the condition -> seems to only accept a variable. Does anyone know how I can create a condition with two db fields with db_select?

Thanks for any help you can bring to me.

Answer 1

Use $ query-> where ($ snippet, $ args = array ());

 $query->where('cfsfc.feature_support_compared_target_id = fpffsfc.feature_support_compared_target_id');