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嘿，目前我们有一个从一个表（关系）中选择所有的数组，我们需要将 Accounts 表中的 account_name 放入该数组中。我们将如何做到这一点？

我们的关系表有（id、receiver_id、sender_id、active、expiry_date）。Receiver_id 和 sender_id 都是 account_id 的外键。

目前它工作正常，但打印了接收者和发送者的 ID，我们希望 Account 表中的 account_name 存在。

这是我们的函数、视图和模型：

功能：

//retrieve Account Id of current User       
        $accountid=$this->Auth->user('account_id');

        //Conditions
        $conditions=array("OR"=> array(
        'Relationship.sender_id' => $accountid,
        'Relationship.receiver_id' => $accountid));

        //Find all Invoices where receiver_id = accountid
        $relationships=$this->Relationship->find('all', array(
        'conditions' => $conditions));

        debug($relationships);

        $this->set('accountid', $accountid); 
        $this->set('relationship', $relationships); 
    }

看法：

<table>
                <tr>
                    <th>Relationship #</th>
                    <th>Sender</th>
                    <th>Receiver</th>
                    <th>Expiry Date</th>
                    <th>Status</th>
                </tr>

        <?php foreach($relationship as $relationships):?>

        <?php

        if($relationships['Relationship']['active']==1)
        {
            $status = 'Active';
        }
        else if($relationships['Relationship']['active']==0)
        {
            $status = 'Not Active';
        }


?>

                    <tr>
                        <td align='center'><?php echo $relationships['Relationship']['id']; ?></td>
                        <td align='center'><?php echo $relationships['Relationship']['sender_id']; ?></td>
                        <td align='center'><?php echo $relationships['Relationship']['receiver_id']; ?></td>
                        <td align='center'><?php echo date('d.m.Y', strtotime($relationships['Relationship']['expiry_date'])); ?></td>
                        <td align='center'><?php echo $status ?></td>
                    </tr>
                <?php endforeach; ?>


            </table>

关系模型：

class Relationship extends AppModel
{

    var $name = 'Relationship';
    public $useTable = 'relationships';
    public $primaryKey = 'id';
    /*
    public $hasMany = array(
        'Invoice' =>
            array(
                'className'              => 'Invoice',
                'joinTable'              => 'invoice',
                'foreignKey'             => 'invoice_id'));*/

    //fix this code
    public $belongsTo = array(
        'User' =>array(
            'className' => 'User',
            'foreignKey' =>'receiver_id',
            'associationForeignKey'  => 'accounts_id',
            )); 

    var $validate = array(
        'date' => array(
            'rule' => array(
                'datevalidation',
                'systemDate'
            ),
            'message' => 'Current Date and System Date is mismatched'
        ),
        'receiver_id' => array(
            'userExists' => array(
                'rule' => array(
                    'userExists',
                ),
                'message' => 'That username doesnt exist.'
            ),
        ),
    );

    function datevalidation($field = array(), $compare_field = null)
    {
        if ($field['date'] > $compare_field)
            return TRUE;
        else
            return FALSE;
    }

    function accountExists($check)
    {   
        $accountExists = $this->Account->find('count', array('conditions' => array('Account.id'=>$check)));
        if ($accountExists == 1) {
           return TRUE;
        }
        else
            return FALSE;
    }

}
4

1 回答 1

0

请注意，它不是来自另一个“控制器”的另一个字段，而是另一个模型。

有多种方法可以做到这一点：

您可以在 find() 方法上使用连接：http: //book.cakephp.org/2.0/en/models/associations-linking-models-together.html#joining-tables

或者您也可以使用 bindModel()，我发现它特别有用：http ://book.cakephp.org/2.0/en/models/associations-linking-models-together.html#creating-and-destroying-associations-即时

搜索食谱！虽然有时很乏味，但它确实非常有帮助。在我对 Cake 的工作原理有一个不错的了解之前，我不得不折腾一番。

于 2012-08-22T14:35:10.033 回答