1

我显然处于深渊。我没有掌握 Doctrine ModelClass 和 Model Table Class 扩展基本模型的好处

例如

class StaffStaff extends Base_StaffStaff
{
    public function getStaffInformation($id_staff_staff){   // == getInformationsByStaff()

        $query = Doctrine_Query::create()
                        ->from("StaffStaff s")
                        ->innerJoin('s.StaffContract c')
                        ->where("s.id_staff_staff = ?", $id_staff_staff);
        $result = $query->execute();

        return $result;

    }
}

并在控制器中

StaffController{

    public function readAction() {

        $id = $this->getRequest()->getParam("id_staff_staff");

        // Get information about a staff
        $model = new StaffStaff();
        $q = $model->getStaffInformation($id);

        $this->view->data = $q[0];

/**
*
* Why do you have to say $q[0] ?
* Is there no better way of doing it?
* How can we access the properties from other tables contained inside the BaseClass extended by the ModelClass
*
*/
}

模型：

/**
 * Base_StaffStaff
 * 
 * This class has been auto-generated by the Doctrine ORM Framework
 * 
 * @property integer $id_staff_staff
 * @property integer $fk_id_staff_individual
 * @property integer $fk_id_staff_team
 * @property integer $fk_id_staff_function
 * 
 */
4

1 回答 1

1

当使用 Query Builder API 时，这是您在getStaffInformation()函数中组装查询的方式，该execute()方法返回一个可迭代游标。这个游标本质上是一个数组，这就是 $q[0] 将访问第一个元素的原因。

如果您尝试只返回一个结果，则应getSingleResult()改用：

$query = Doctrine_Query::create()
   ->from("StaffStaff s")
   ->innerJoin('s.StaffContract c')
   ->where("s.id_staff_staff = ?", $id_staff_staff);
return $query->getSingleResult();

另一方面，可迭代光标很整洁，因为您可以执行以下操作：

$staff = $model->getStaffInformation($id);
foreach($staff as $person)
{
   // Assign all staff to view array
   $this->view->staff[] = $person;
}
于 2012-06-16T01:08:27.867 回答