【问题标题】:stuck with zend_db_table + join卡在 zend_db_table + join
【发布时间】:2011-07-25 09:10:55
【问题描述】:

试图理解使用Zend_Db_Table

我有一张这样的桌子:

然后我创建了类:

class table_1 extends Zend_Db_Table_Abstract
{
    protected $_name = 'table_1';
    protected $_primary = 't1_id';
    protected $_referenceMap    = array(
        'DepCard' => array(
            'columns'           => 't1_id',
            'refTableClass'     => 'table_2',
            'refColumns'        => 't2_t1'
        ),
        'Select1' => array(
            'columns' => array('t1_select1'),
            'refTableClass' => 'Select_1'
        ),
        'Select2' => array(
            'columns' => array('t1_select2'),
            'refTableClass' => 'Select_2'
        )
    );

}

class table_2 extends Zend_Db_Table_Abstract {
  protected $_dependentTables = array('table_1');
}
class Select_1 extends Zend_Db_Table_Abstract {
  protected $_dependentTables = array('table_1');
}
class Select_2 extends Zend_Db_Table_Abstract{
  protected $_dependentTables = array('table_1');
}

那我想:

$table_1 = new table_1();
$data = $table_1->fetchAll();

与所有从属表。 有没有办法做到这一点?

【问题讨论】:

    标签: zend-db-table zend-db-select


    【解决方案1】:

    查看Zend_Db_TableZend_Db_Table_Select 的代码表明这是不可能的。它始终只在单个表上运行fetchAll

    【讨论】:

    • 所以,唯一的方法是制作class my_table_abstract extends Zend_Db_Table_Abstract 并重载基本功能?可能有简单的方法吗?
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