【问题标题】:Cakephp's beforeSave() not retaining new data using $model::save()Cakephp 的 beforeSave() 不使用 $model::save() 保留新数据
【发布时间】:2014-05-20 12:30:08
【问题描述】:

当我在 SalasController reservar() 方法上执行 debug($this->data) 时,来自 json 的 $data 是可以的,但是当我执行 this->Sala->save($this->data) 并在 beforeSave() 上尝试 debug($data) 时,返回:

Notice (8): Undefined property: FormataDataBehavior::$data [APP\Model\Behavior\FormataDataBehavior.php, line 52]
\app\Model\Behavior\FormataDataBehavior.php (line 52)
null

SalasController.php

App::uses('AppController', 'Controller');

class SalasController extends AppController {

public $components = array('RequestHandler');

public function beforeFilter() {
    parent::beforeFilter();
    $this->Auth->allow('listar', 'reservar');
    $this->RequestHandler->addInputType('json', array('json_decode', true));
}
    public function reservar() {
    $this->Sala->save($this->data); //no conditions just for debugging.
}

模型 Sala.php

class Sala extends AppModel {

public $useTable = 'agenda';
public $actsAs = array(
    'FormataData' => array('dia')
);

行为

class FormataDataBehavior extends ModelBehavior {

// Armazena os campos do model a serem formatados.
public $campos;

//Inicializa o behavior
public function setup(Model $model, $settings = array()) {
    if (!empty($settings)) {
        $this->campos[$model->name] = $settings;
    } else {
        throw new MethodNotAllowedException('Campos data não informados na declaração do behavior');
    }
}

public function beforeSave(Model $model, $options = array()) {
    parent::beforeSave($model, $options);
    debug($this->data); //trying show data, but is !isset... =/
    die();
}

MyJs

function adicionarReserva() {
var dadosForm = {Sala: {}};
$.each($("#form-reservas").serializeArray(), function(index, value) {
    dadosForm.Sala[value.name] = value.value;
});
$.post("/intracake/Salas/reservar.json", dadosForm)
        .done(function(data) {
            sysMsg(data.html);
            atualizaTabelas();
        });
}

【问题讨论】:

  • FormataDataBehavior 是一种行为,而不是模型,所以我认为你应该使用$model->data 而不是$this->data
  • $data = $model->read(); ?
  • $model->data(我编辑了我的评论),data 是根据 CakePHP api 的 Model 的公共属性,因此您可以从任何地方访问它。
  • 所以,我会发布一个答案......你可以检查是否是一个好习惯?我发布代码。
  • 因为您查看的是 Model 类的文档,而不是 ModelBehaviour 类的文档。

标签: php ajax cakephp callback requesthandler


【解决方案1】:

@Holt 提示后我的 beforeSave;

public function beforeSave(Model $model, $options = array()) {
    parent::beforeSave($model, $options);
    $this->formataBeforeSave($model);
    return true;
}

public function formataBeforeSave(Model &$model, $padrao = 'Y-m-d') {
    foreach ($this->campos[$model->name] as $campo) {
        if (isset($model->data[$model->name][$campo]) && !empty($model->data[$model->name][$campo])) {
            $model->data[$model->name][$campo] = $this->converte($model, $model->data[$model->name][$campo], $padrao);
        }
    }
}

【讨论】:

    猜你喜欢
    • 2011-06-14
    • 1970-01-01
    • 1970-01-01
    • 2023-03-15
    • 2011-08-08
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 2020-02-06
    相关资源
    最近更新 更多