Phalcon:如何得到验证有关目前保存？

Question

我有一个型号的音频和一个模型AudioCategory.
当我保存的声音目的，我想证实，至少有1audiocategory。

我创建了一个自定义验证程序。
我试图使用元音->getRelated()在验证，但它一直试图获取数据库中的信息。由于验证之前发生的节(这是伟大的)，然后我收到一个空的名单，因此我的验证程序总是返回错误的。

当我打印的声音对象，而不保存，我可以看到我audiocategory领域_related的音频对象(print_r($音频);):
[_related：保护]=>阵列
( [audiocategory]=>阵列
(
[0]=>GRQ\音频\AudioCategory Object([...])
[1]=>GRQ\音频\AudioCategory Object([...])
)
)

如果我尝试print$音频->audiocategory直接，我得到通知：
访问不确定性GRQ\音频\音::audiocategory
并没有什么是返回。

如果我呼叫$音频->getRelated(),我得到的一个目的类型Phalcon\软\型\结果集\简单的一个空_result.(这是逻辑的，因为它去搜索的数据库...)

---

因此，我的问题是：
我如何可以得到并核实有关领域之前拯救他们吗？

---

这里是我的(缩短)控制器试验：

    $audioCategory = new AudioCategory();
    $audioCategory->categoryId = 1;
    $arAudioCategory[0] = $audioCategory; 

    $audioCategory = new AudioCategory();
    $audioCategory->categoryId = 2;
    $arAudioCategory[1] = $audioCategory;

    $audio = new Audio();
    [...other fields initialization...]
    $audio->audiocategory = $arAudioCategory;
    $audio->save();

这里是(缩短的)音模型：

namespace GRQ\Audio;
use GRQ\Validator\PresenceOfRelationValidator;
class Audio extends \Phalcon\Mvc\Model {
/**
 * @Primary
 * @Identity
 * @Column(type="integer", nullable=false)
 */
public $id = 0; 
/**
 * @Column(type="integer", nullable=false)
 */
public $createdAt = 0;

[...other fields all reflecting the database...]

public function initialize() {
    $this->setSource ( "audio" );

    // table relationships
    $this->hasMany ( "id", "GRQ\Audio\AudioCategory", "audioId", array(
            'alias' => 'audiocategory'
    ) );
}

public function validation() {      
    [...other validations...]

    $this->validate ( new PresenceOfRelationValidator ( array (
            "field" => "audiocategory" 
    ) ) );

    return $this->validationHasFailed () != true;
}
}

这里是(缩短)的音频类型：

namespace GRQ\Audio;    
class AudioCategory extends \Phalcon\Mvc\Model {
/**
 * @Primary
 * @Identity
 * @Column(type="integer", nullable=false)
 */
public $id = 0; 
/**
 * @Column(type="integer", nullable=false)
 */
public $audioId = 0;    
/**
 * @Column(type="integer", nullable=false)
 */
public $categoryId = 0;

public function initialize(){
    $this->setSource("audiocategory");
    //table relationships
    $this->belongsTo("audioId", "GRQ\Audio\Audio", "id", array(
            'alias' => 'audio'
    ));
}
}

这是我的定义验证程序(其中不工作，并总是返回false):

namespace GRQ\Validator;

use Phalcon\Mvc\Model\Validator;
use Phalcon\Mvc\Model\ValidatorInterface;

class PresenceOfRelationValidator extends Validator implements ValidatorInterface {
public function validate($model){
    $field = $this->getOption('field');
    $message = $this->getOption('message');
    if (!$message) {
        $message = 'The required relation '.$field.' was not found';
    }

    $value = $model->getRelated($field);

    if (count($value) == 0) {
        $this->appendMessage(
                $message,
                $field,
                "PresenceOfRelation"
        );
        return false;
    }
    return true;
}
}

Answer 1

所以,我找到一种方式实现这一目标。不确定这是最好的方式，但它的工作：
由于价值得到保护，我不得不让他们从我的对象。
所以我创造了一个基本模型，从而延长自己说：

基本模型：

namespace GRQ;
class BaseModel extends \Phalcon\Mvc\Model {

/**
 * This function should be used to get the data in the _related field directly.
 * It is very useful if you need to validate the presence of a relation BEFORE saving in the database.
 * To initialize the field with the database content, use $this->getRelated().
 */
public function getInternalRelated(){
    return $this->_related;
}   
}

然后我改变了我的音频类延长，从我的基本模型：

音频模型(简化):

namespace GRQ\Audio;

use Phalcon\Mvc\Model\Validator\Numericality;
use GRQ\Validator\MinValueValidator;
use GRQ\Validator\PresenceOfRelationValidator;

class Audio extends \GRQ\BaseModel {
/**
 * @Primary
 * @Identity
 * @Column(type="integer", nullable=false)
 */
public $id = 0;

/**
 * @Column(type="string", length=255, nullable=false)
 */
public $title = '';

public function initialize() {
    $this->setSource ( "audio" );

    // table relationships
    $this->hasMany ( "id", "GRQ\Audio\AudioCategory", "audioId", array(
            'alias' => 'audiocategory'
    ) );
}

public function validation() {              
    $this->validate ( new PresenceOfRelationValidator ( array (
            "field" => "audiocategory" 
    ) ) );

    return $this->validationHasFailed () != true;
}
}

我AudioCategory模型(简化)留下的差不多相同的：

namespace GRQ\Audio;

use Phalcon\Mvc\Model\Message;

class AudioCategory extends \GRQ\BaseModel {
/**
 * @Primary
 * @Identity
 * @Column(type="integer", nullable=false)
 */
public $id = 0;

/**
 * @Column(type="integer", nullable=false)
 */
public $audioId = 0;

/**
 * @Column(type="integer", nullable=false)
 */
public $categoryId = 0;

public function initialize()
{
    $this->setSource("audiocategory");
    //table relationships
    $this->belongsTo("audioId", "GRQ\Audio\Audio", "id", array(
            'alias' => 'audio'
    ));
}
}

我现在使用验证程序的方法getInternalRelated到验证：

namespace GRQ\Validator;

use Phalcon\Mvc\Model\Validator;
use Phalcon\Mvc\Model\ValidatorInterface;

class PresenceOfRelationValidator extends Validator implements ValidatorInterface {
public function validate($model){
    $field = $this->getOption('field');
    $message = $this->getOption('message');
    if (!$message) {
        $message = 'The required relation '.$field.' was not found';
    }

    $value = $model->getInternalRelated();

    if (count($value[$field]) == 0) {
        $this->appendMessage(
                $message,
                $field,
                "PresenceOfRelation"
        );
        return false;
    }
    return true;
}
}