Question

假设有两个表。

Table X--
Columns:
     id         x_value

Table Y--
Columns:
     id        x_id      y_value

现在我不想在doctrine类中定义关系，我想使用这样的查询使用这两个表检索一些记录：

Select x_value from x, y where y.id="variable_z" and x.id=y.x_id;

我无法弄清楚如何在doctrine orm

中编写这样的查询

修改

表格结构：

表1：

CREATE TABLE IF NOT EXISTS `image` (
  `id` int(11) NOT NULL AUTO_INCREMENT, 
`random_name` varchar(255) NOT NULL,
 `user_id` int(11) NOT NULL,
  `community_id` int(11) NOT NULL,
  `published` varchar(1) NOT NULL,
  PRIMARY KEY (`id`)
 ) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=259 ;

表2：

  CREATE TABLE IF NOT EXISTS `users` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
   `city` varchar(20) DEFAULT NULL,
   `state` varchar(20) DEFAULT NULL,
   `school` varchar(50) DEFAULT NULL,
   PRIMARY KEY (`id`)
 ) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ;

我正在使用的查询：

        $q = new Doctrine_RawSql();

        $q  ->select('{u.*}, {img.*}')
        ->from('users u LEFT JOIN image img ON u.id = img.user_id')
        ->addComponent('u', 'Users u')
        ->addComponent('img', 'u.Image img')
        ->where("img.community_id='$community_id' AND img.published='y' AND u.state='$state' AND u.city='$city
        ->orderBy('img.id DESC')
        ->limit($count+12)  
        ->execute();

我收到错误：

 Fatal error: Uncaught exception 'Doctrine_Exception' with message 'Couldn't find class
 u' in C:\xampp\htdocs\fanyer\doctrine\lib\Doctrine\Table.php:290 Stack trace: #0 
 C:\xampp\htdocs\fanyer\doctrine\lib\Doctrine\Table.php(240): Doctrine_Table-   >initDefinition() #1 C:\xampp\htdocs\fanyer\doctrine\lib\Doctrine\Connection.php(1127): 
Doctrine_Table->__construct('u', Object(Doctrine_Connection_Mysql), true) #2 
C:\xampp\htdocs\fanyer\doctrine\lib\Doctrine\RawSql.php(425): Doctrine_Connection-
>getTable('u') #3 C:\xampp\htdocs\fanyer\doctrine\models\Image.php(33):    Doctrine_RawSql-
>addComponent('img', 'u.Image imga') #4   C:\xampp\htdocs\fanyer\community_images.php(31): 
  Image->get_community_images_gallery_filter(4, 0, 'AL', 'ALBERTVILLE') #5 {main} thrown      in 
    C:\xampp\htdocs\fanyer\doctrine\lib\Doctrine\Table.php  on line 290

Answer 1

尝试这样的事情：---

$q = new Doctrine_RawSql();
$this->related_objects = $q->
        select('{o.name}')->
        from('tagset t1 JOIN tagset t2 ON t1.tag_id = t2.tag_id AND t1.object_id != t2.object_id JOIN object o ON t2.object_id = o.id')->
        addComponent('o','Object o')->
        where('t1.object_id = ?', $this->object->id)->
        groupBy('t2.object_id')->
        orderBy('COUNT(*) DESC')->
        execute();

Answer 2

您可以将sql直接写入db驱动程序。但是，您不会返回任何水合数组或对象。但它有时会派上用场：

 $sql = "SELECT * FROM... ";        
 $pdo = Doctrine_Manager::connection()->getDbh();       
 $data = $pdo->query($sql)->fetchAll();

Answer 3

您可以在Symfony 1.4中手动更改表类，但这应该仔细完成，因为您没有链接通常应该发生的数据库级别的表。

class TableName extends BaseTableName
{
    public function setUp()
    {
        $this->hasOne('LinkedTableName', array(
             'local' => 'link_id',
             'foreign' => 'link_id'));
        parent::setUp();
    }
}

可能重复：doctrine join without relation

是否可以在不使用关系的情况下连接doctrine ORM中的表？

3 个答案: