Question

我正在将这个查询应用于D6以下的查询，而不是工作..不知道我错了我做....内部连接在某些条件下失败

$result = db_select('px_slides','s')
    ->join('node','n','s.vid = n.vid')
    ->fields('s',array('tissue_type','body_site'))
    ->fields('n',array('sticky','title'))
    ->condition('n.status','1','=')
    ->condition('s.cid','126','=')
    ->execute()->fetchObject();

drupal 6查询我有：

    $result = db_query('
  SELECT n.nid, n.vid, n.sticky, n.title, n.created, s.cid, s.ref_id, s.viewurl, s.specimen_type, s.tissue_type, s.body_site, s.test_type, s.algorithm, s.result 
  FROM {px_slides} s INNER JOIN {node} n ON n.vid = s.vid
  WHERE n.status = 1 ')->execute();

Answer 1

您需要将->join()的呼叫完全放在一个单独的行上，因为它不会返回查询对象：

$query = db_select('px_slides','s')  
  ->fields('s',array('tissue_type','body_site'))
  ->fields('n',array('sticky','title'))
  ->condition('n.status','1','=')
  ->condition('s.cid','126','=');

$query->join('node','n','s.vid = n.vid');

$result = $query->execute()->fetchObject();

Answer 2

join方法不会像那样链接。你必须做类似的事情：

$query = db_select('px_slides','s')
    ->join('node','n','s.vid = n.vid');
$query->fields('s',array('tissue_type','body_site'))
    ->fields('n',array('sticky','title'))
    ->condition('n.status','1','=')
    ->condition('s.cid','126','=');
$result = $query->execute()->fetchObject();

此外，您可以使用to string magic方法查看它将要执行的查询。

$query->__toString();

Answer 3

试试这个......

$query = db_select('px_slides', 's');
$query->innerJoin('node,'n','s.vid = n.vid');
$query->fields('s',array('tissue_type','body_site'));
$query->fields('n',array('sticky','title'));
$query->condition('n.status','1');
$query->condition('s.cid','126');
$result= $query->execute()->fetchAll(PDO::FETCH_ASSOC);

drupal 7中的内连接查询

3 个答案: