小编n4t*_*han的帖子

我一直在使用cakephp paginations选项2天.我需要创建一个INNER联接列出几个字段,但我必须处理搜索以过滤结果.这是我处理搜索选项的代码的一部分$this->passedArgs

function crediti() {

    if(isset($this->passedArgs['Search.cognome'])) {
                debug($this->passedArgs);

                $this->paginate['conditions'][]['Member.cognome LIKE'] = str_replace('*','%',$this->passedArgs['Search.cognome']);

        }
        if(isset($this->passedArgs['Search.nome'])) {
                $this->paginate['conditions'][]['Member.nome LIKE'] = str_replace('*','%',$this->passedArgs['Search.nome']);

        }

之后

$this->paginate = array(

            'joins' => array(array('table'=> 'reservations',
            'type' => 'INNER',
            'alias' => 'Reservation',
            'conditions' => array('Reservation.member_id = Member.id','Member.totcrediti > 0' ))),
            'limit' => 10);
        $this->Member->recursive = -1;
        $this->paginate['conditions'][]['Reservation.pagamento_verificato'] = 'SI';
        $this->paginate['fields'] = array('DISTINCT Member.id','Member.nome','Member.cognome','Member.totcrediti');
        $members = $this->paginate('Member');
        $this->set(compact('members'));

INNER JOIN工作得很好,但$ this-> paginations忽略了每一个$this->paginate['conditions'][] by $this->passedArgs,我不知道如何解决它.调试中没有查询,只是原始查询INNER JOIN.有人可以帮助我吗？非常感谢你

更新:没有运气.我已经处理了这部分代码很长时间了.如果我使用

if(isset($this->passedArgs['Search.cognome'])) {
                    $this->paginate['conditions'][]['Member.cognome LIKE'] = str_replace('*','%',$this->passedArgs['Search.cognome']);

            }
$this->paginate['conditions'][]['Member.sospeso'] …