是否有更优化和更短的方式来获得具有特定条件的节点?EntityFieldQuery多个替代条件
$query = new EntityFieldQuery;
$result = $query
->entityCondition('entity_type', 'node')
->propertyCondition('type', $node_type)
->propertyCondition('title', $title)
->fieldCondition('field_number', 'value', '1', '=');
->propertyCondition('status', 1, '=')
->execute();
// $result['node'] contains a list of nids where the title matches
if (!empty($result['node']) {
// You could use node_load_multiple() instead of entity_load() for nodes
$nodes = entity_load('node', array_keys($result['node']));
}
$query_two = new EntityFieldQuery;
$result_two = $query_two
->entityCondition('entity_type', 'node')
->propertyCondition('type', $node_type)
->propertyCondition('title', $title)
->fieldCondition('field_number', 'value', '2', '=');
->propertyCondition('status', 1, '=')
->execute();
// $result_two['node'] contains a list of nids where the title matches
if (!empty($result_two['node']) {
// You could use node_load_multiple() instead of entity_load() for nodes
$nodes_two = entity_load('node', array_keys($result_two['node']));
}
感谢您花时间回答这个问题并感谢EFQ,这是一个很棒的功能。 – Sam152 2012-06-28 01:40:08