CakePHP 2 中的表单验证

Question

在我看来我有

<?= $this->Form->create('Asc201516', array(
'url' => array(
    'controller' => 'asc201516s', 
    'action'     => 'submit_asc201516'
    ), 
'class' => 'form-inline',
'onsubmit' => 'return check_minteacher()'
)); ?>

<div class="form-group col-md-3">
    <?= $this->Form->input('bi_school_na', array(
        'type' => 'text',
        'onkeypress' => 'return isNumberKey(event)',
        'label'       => 'NA', 
        'placeholder' => 'NA', 
        'class'       => 'form-control'
    )); ?>
</div>
<?php 
$options = array(
    'label' => 'Submit',
    'class' => 'btn btn-primary');
echo $this->Form->end($options); 
?>

在我的控制器中，我有

$this->Asc201516->set($this->request->data['Asc201516']);
        if ($this->Asc201516->validates()) {
            echo 'it validated logic';
            exit();

        } else {                
            $this->redirect(
                array(
                    'controller' => 'asc201516s',
                    'action' => 'add', $semisid
                    )
                );

        }

在我的模型中，我有

 public $validate = array(
        'bi_school_na' => array(
            'Numeric' => array(
                'rule' => 'Numeric',
                'required' => true,
                'message' => 'numbers only',
                'allowEmpty' => false
            )
        )
    );

当我提交表单时，从逻辑上讲，它不应该被提交并打印出错误消息，而是提交了表单并验证了控制器内的模型，这会破坏控制器中的操作。

Answer 1

您必须在您的控制器中检查验证，例如

$this->Asc201516->set($this->request->data);  
if($this->Asc201516->validates()){  
    $this->Asc201516->save($this->request->data);  
}else{  
    $this->set("semisid",$semisid);
    $this->render("Asc201516s/add");
}

您将在变量 $semisid 中拥有您的 ID，或者您可以在 $this->request->data = $this->Asc201516->findById($semisid);

中设置数据