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Php Zend\u Db\u表\u选择\u异常_Php_Zend Framework_Zend Db_Zend Db Table - Fatal编程技术网

Php Zend\u Db\u表\u选择\u异常

Php Zend\u Db\u表\u选择\u异常,php,zend-framework,zend-db,zend-db-table,Php,Zend Framework,Zend Db,Zend Db Table,是否从以下方法收到此错误 2013-10-02T14:20:05+01:00错误(3):异常“Zend_Db_Table_Select_exception”,在/usr/share/php/Zend/Db/Table/Select.php:215中显示消息“Select query cannot join with other Table” class Application_Model_DbTable_Provenir extends Zend_Db_Table_Abstract {

是否从以下方法收到此错误

2013-10-02T14:20:05+01:00错误(3):异常“Zend_Db_Table_Select_exception”,在/usr/share/php/Zend/Db/Table/Select.php:215中显示消息“Select query cannot join with other Table”

class Application_Model_DbTable_Provenir extends Zend_Db_Table_Abstract {
    protected $_name = 'provenir_instance_response';
    protected $_primary = 'provenir_response_id';
    protected $_sequence = true;

    /**
    * Get passed result and check ID
    * @param string $memId
    * @return Zend_Db_Table
    */
    public function scoreCardQuery($memId){
        //Build Query
        $select = $this->select();
        $select->from('scorecard_results', array('passed', 'check_id'));
        $select->where('traveller_id=?', $memId);
        return $this->fetchAll($select);
    }
}
那里没有会合吗

是因为我引用了
provenir\u instance\u response
作为类参数中的DB表,但我正在尝试从
scorecard\u结果中进行选择吗

/**
 * Get passed result and check ID
 * @param string $memId
 * @return Zend_Db_Table
 */
public function scoreCardQuery($memId){
    //Build Query
    $select = $this->select();
    $select->setIntegrityCheck(false); //Must be set for this query to work
    $select->from('scorecard_results', array('passed', 'check_id'));
    $select->where('traveller_id=?', $memId);
    return $this->fetchAll($select);
}