Cakephp $ this->使用自定义JOIN和过滤选项进行分页

Question

我已经使用了Cakephp分页选项2天了。我需要进行“内部联接”以列出一些字段，但是我必须处理搜索以过滤结果。
这是代码的一部分，其中我通过$this->passedArgs处理搜索选项

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的效果很好，但是$ this-> paginations忽略了每个 $this->paginate['conditions'][] by $this->passedArgs，我不知道该如何解决。
在调试中没有查询，只有原始的 INNER JOIN。
有人可以帮我吗？
非常感谢你

更新:
没有运气。
我已经处理了这部分代码很多小时了。
如果我用

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

            }
$this->paginate['conditions'][]['Member.sospeso'] = 'SI';
        $this->Member->recursive = 0;
        $this->paginate['fields'] = array(
            'Member.id','Member.nome','Member.cognome','Member.codice_fiscale','Member.sesso','Member.region_id',
            'Member.district_id','Member.city_id','Member.date','Member.sospeso','Region.name','District.name','City.name');
        $sospesi = $this->paginate('Member');

一切顺利，从调试中我收到了第一个条件，以及 $this->paginate['conditions'][]['Member.cognome LIKE']的条件，如您所见
数组 $this->passedArgs

Array
(
    [Search.cognome] => aiello
)

Array $this->paginate['conditions'][]
(
    [0] => Array
        (
            [Member.cognome LIKE] => aiello
        )

    [1] => Array
        (
            [Member.sospeso] => NO
        )

但是，如果我用paginate编写联接， $this->paginate['conditions'][]将忽略所有内容，并从调试中给我，只是 $this->paginate['conditions'][]['Reservation.pagamento_verificato'] = 'SI';另一点信息。
如果我把所有与 $this->paginate['conditions'][]['Reservation.pagamento_verificato'] = 'SI';有关的东西放进去
在 $this->paginate JOIN之前， $this->paginate['conditions'][]中将没有任何内容。

Answer 1

这是一个古老的问题，因此，我将回顾一下如何像我一样从谷歌到这里的其他人分页进行JOIN。这是来自窗口小部件 Controller 的示例代码，将Widget.user_id FK连接到User.id列，仅显示当前用户(在条件下):

// Limit widgets shown to only those owned by the user.
$this->paginate = array(
    'conditions' => array('User.id' => $this->Auth->user('id')),
    'joins' => array(
        array(
            'alias' => 'User',
            'table' => 'users',
            'type' => 'INNER',
            'conditions' => '`User`.`id` = `Widget`.`user_id`'
        )
    ),
    'limit' => 20,
    'order' => array(
        'created' => 'desc'
    )
);
$this->set( 'widgets', $this->paginate( $this->Widget ) );

这使得查询类似于:

SELECT widgets.* FROM widgets 
INNER JOIN users ON widgets.user_id = users.id 
WHERE users.id = {current user id}

并且仍然分页。