Question

我有一个模型Audio和一个模型AudioCategory 当我保存音频对象时，我想验证是否附加了至少1个audiocategory。

我为此创建了一个自定义验证器我尝试在验证器中使用$ audio-＆gt; getRelated（），但它一直试图在数据库中获取信息。由于验证发生在保存之前（这很好），然后我收到一个空列表，因此我的验证器总是返回false。

当我打印音频对象而不保存时，我可以在声音对象的字段_（print_r（$ audio）;）中看到我的audiocategory：） [_related：protected] =＆gt;阵列
    （         [audiocategory] =＆gt;阵列
                （
                    [0] =＆gt; GRQ \ Audio \ AudioCategory对象（[...]）
                    [1] =＆gt; GRQ \ Audio \ AudioCategory对象（[...]）
                ）
        ）

如果我尝试直接打印$ audio-＆gt; audiocategory，我会收到通知：
访问未定义的属性GRQ \ Audio \ Audio :: audiocategory
没有任何回报。

如果我调用$ audio-＆gt; getRelated（），我会得到一个Phalcon \ Mvc \ Model \ Resultset \ Simple类型的对象，其中包含空的_result。（这是逻辑，因为它在数据库中进行搜索......）

因此，我的问题是：
如何在保存之前获取并验证相关字段？

这是我的（缩短的）控制器测试：

    $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();

这是（缩短的）音频模型：

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;
}
}

以下是（缩短的）音频类别模型：

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'
    ));
}
}

这是我的自定义验证器（它不起作用并始终返回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，我找到了实现这一目标的方法。不确定这是最好的方式，但它有效：
由于值受到保护，我不得不将它们从我的对象中暴露出来所以我创建了一个基础模型来扩展自己：

基本型号：

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;
}   
}

然后我将我的音频类改为从我的基本模型扩展：

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;
}
}

我的AudioCategory模型（简化）几乎保持不变：

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'
    ));
}
}

我的Validator现在使用方法getInternalRelated来验证：

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;
}
}

Phalcon：如何在保存之前获取/验证相关对象？

1 个答案: