CakePHP finds elements without writing HABTM relationships
Simply put: Tags HABTM Documents
Is there a way to find all Tags
unrelated Document
?
I started with this:
$freeTags = $this->Tag->find('all', array(
'conditions' => array(
),
'contain' => array(
'Document'
),
'recursive' => -1
))
but I have no idea how to get a request like this:
SELECT * FROM tags WHERE id NOT IN (SELECT tag_id FROM documents_tags)
My CakePHP version is 2.3
EDIT: Final solution, method in Tag
model
$db = $this->getDataSource();
$subQuery = $db->buildStatement(
array(
'fields' => array('DocumentsTag.tag_id'),
'table' => $db->fullTableName($this->DocumentsTag),
'alias' => 'DocumentsTag',
'limit' => null,
'offset' => null,
'joins' => array(),
'conditions' => null,
'order' => null,
'group' => null
), $this->DocumentsTag
);
$subQuery = ' Tag.id NOT IN (' . $subQuery . ') ';
$subQueryExpression = $db->expression($subQuery);
$conditions[] = $subQueryExpression;
$freeTags = $this->find('all', compact('conditions'));
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1 answer
How CookBook :: Sub queries you could make
$db = $this->User->getDataSource();
$subQuery = $db->buildStatement(
array(
'fields' => array('"DocumentsTag"."tag_id"'),
'table' => $db->fullTableName($this->DocumentsTag),
'alias' => 'DocumentsTag',
'limit' => null,
'offset' => null,
'joins' => array(),
'conditions' => null,
'order' => null,
'group' => null
),
$this->DocumentsTag
);
$subQuery = ' "Tag"."id" NOT IN (' . $subQuery . ') ';
$subQueryExpression = $db->expression($subQuery);
$conditions[] = $subQueryExpression;
$this->User->find('all', compact('conditions'));
I hope this can be helpful for you.
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