Question

我是CakePHP的新手，我正在阅读手册并买了这本书，我遇到了一个问题，需要一些帮助。

目前我这样查询我的数据库（这是全班）：

<?php

class TapplicantController extends AppController {

    public function index() {

        $this->set('page_title', 'Dashboard');

        $tapplicant = $this->Tapplicant->query("SELECT tapplicant.AppID, tapplicant.AppIPAddress, tapplicant.AppAffID, tapplicant.AppDate, tapplicant.FirstName, tapplicant.LastName, tapplicant.Email, tapplicant.AppDomain, toutcome.AffCommission FROM tapplicant INNER JOIN toutcome ON tapplicant.AppID=toutcome.AppID LIMIT 0,15 ");

        $this->set('tapplicant', $tapplicant );     


    }
}

然后我像这样显示它并且工作正常：

<?php foreach ($tapplicant as $tapplicant) : ?>
        <tr>

            <td><?=$tapplicant['tapplicant']['AppID']; ?></td> 
            <td><?=$tapplicant['tapplicant']['AppAffID']; ?></td> 
            <td><?=$tapplicant['tapplicant']['FirstName']; ?></td>
            <td><?=$tapplicant['tapplicant']['LastName']; ?></td>
            <td><?=$tapplicant['tapplicant']['Email']; ?></td>
            <td><?=$tapplicant['tapplicant']['AppDate']; ?></td>
            <td><?=$tapplicant['tapplicant']['AppDomain']; ?></td>
            <td><?=$tapplicant['tapplicant']['AffCommission']; ?></td>

        </tr>
        <?php endforeach; ?>

问题是我有Inner Joined 2表并且它不起作用，我收到以下错误：未定义的索引：AffCommission

Answer 1

除非您真的无法使用Cake way编写查询，否则您不应该使用query()来检索CakePHP中的数据。要在Cake中使用连接，您的查询应如下所示： -

$tapplicant = $this->Tapplicant->find(
    'all',
    array(
        'fields' => array(
            'Tapplicant.*',
            'Toutcome.*'
        ),
        'joins' => array(
            array(
                'table' => 'toutcome',
                'alias' => 'Toutcome',
                'type' => 'INNER',
                'conditions' => array('Tapplicant.AppID = Toutcome.AppID')
            )
        ),
        'limit' => 15
    )
);

然后，应该通过模型别名索引生成的数组，因此$tapplicant['tapplicant']['AffCommission'];应该类似于$tapplicant['Toutcome']['AffCommission'];。调试此方法的最简单方法是查看返回的数组是什么，因此在代码中使用print_r($tapplicant)或debug($tapplicant)来查看返回的内容。

只要您正确地关联模型，Containable behaviour可能会使它更简单。

你没有使用CakePHP naming conventions你真的应该这样做，否则你在使用Cake时会遇到很多困难。我建议您从一开始就阅读在线文档，然后再深入了解！

Answer 2

Try:
$options['joins'] = array(
    array(
        'table' => 'toutcome',
        'alias' => 'Toutcome',
        'type' => 'INNER',
        'conditions' => array('Tapplicant.AppID = Toutcome.AppID')
         )
);

$options['fields'] = array(
    Tapplicant.AppID, 
    Tapplicant.AppIPAddress, 
    Tapplicant.AppAffID, 
    Tapplicant.AppDate, 
    Tapplicant.FirstName, 
    Tapplicant.LastName, 
    Tapplicant.Email, 
    Tapplicant.AppDomain, 
    Toutcome.AffCommission
);

$options['limit'] = 15;

$tapplicant = $this->Tapplicant->find('all', $options);

cakePHP中的mySQL Join Query

2 个答案: