相关疑难解决方法(0)

发生错误应用程序错误异常信息:

消息:选择查询无法与另一个表连接这是我的代码:

<?php

class Application_Model_DbTable_Seguimientos extends Zend_Db_Table_Abstract {

    protected $_name = 'Seguimientos';    
    public function buscarCaso($cod_beneficiario) {
        $consulta = $this->select()

        ->from(array('seg' => 'Seguimientos'))
        ->join(array('casos' => 'Asuntos_Estudiantiles'),
        'seg.cod_radicado = casos.codigo_radicado')
        ->where('casos.cod_beneficiario = ?', $cod_beneficiario);

        $query = $this->fetchAll($consulta)->toArray();
        return $query;
    }
}

我使用Zend Framework 1

我在zend中使用Join查询..喜欢

$select = $table->select()
                         ->from(array('e' => 'EducationHistory'),
                                array('status_DataDictionary_id'))
                             ->join(array('r' => 'ReportOrder'),
                                    'e.id = r.EducationHistory_id',
                                    array('reportOrderStatusId' => 'r.status_DataDictionary_id'))
                        ->where('r.orderBy_Organization_id = ?', 4) 
                        ->where('r.orderBy_Person_id = ?', 1)            
                        ->group('e.enrollno');

为此,我从http://framework.zend.com/manual/en/zend.db.select.html获取帮助

但是当我尝试运行该查询时,会出现一个错误,说明了这一点

选择查询无法与其他人联接

任何人都可以帮助我吗？提前致谢.... :)

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