Question

now, if the node type is article. on the term page’s left, i want to show the latest 10 articles’s title which node type is article. i don’t want to use views, how do i do? thank you.

if i want to show the latest 10 articles’s title which node type is article on the node page’s left . how to write the query. many thanks.

ps:i found EntityFieldQuery maybe can do this, but i don’t now how to do out put it.

my code:

$query = new EntityFieldQuery();

$query
 ->entityCondition('entity_type', 'node')
 ->entityCondition('bundle', 'article')
 ->propertyCondition('status', 1)
 ->propertyOrderBy('created', 'DESC')
  ->range(0, 10);

$result = $query->execute();

Answer 1

The code can be something like that (using db_select())

$query = db_select("node", "n") // select from the node table
    ->fields("n", array("nid", "title")) // fields nid, title
    ->condition("type", "page", "=") // where the node type = page
    ->orderBy("created", "DESC") // order by the newest
    ->range(0, 10) // select only 10 records
    ->execute(); // execute the query

while($record = $query->fetchAssoc()) {
    print(l($record['title'], "node/" . $record['nid'])); // print the node title linked to node.
}

Another example using EntityFieldQuery():

$query = new EntityFieldQuery();
$entities = $query->entityCondition('entity_type', 'node')          
      ->entityCondition('bundle', 'club')
      ->propertyOrderBy("created", "DESC")
      ->range(0, 10)
      ->execute();

foreach($entities['node'] as $obj)
{
    $node = node_load($obj->nid);
    print(l($node->title, "node/" . $node->nid));
}

Performance wise: use the first method.