Question

我正在使用Cakephp 2，我确实在动作中生成了html内容我需要将html内容发送到空白视图。

我有控制器动作发送：

public function send()  {
        if ($this->request->is('get')) {
           $this->request->data['Doc']['value'] = 0.01;        
           $this->request->data['Doc']['name'] = 'OneName';
        }        
        if ($this->request->is('post')) {              
            echo $this->generate($this->request->data);              
        }       
    }

行动产生：

public function generate($data)   {    
    $html= new htmlGenerator($data);
    return $html->getOutput();
}

我有观看send.ctp和generate.ctp

当我发送表单时，生成的html会保留在当前视图的顶部

如何发送到generate.ctp视图，传递this-＆gt; request-＆gt;数据以生成操作？

Answer 1

这应该是这样的

public function send()  {       
    $this->autoRender = false;      

    if ($this->request->is('get')) {
       $this->request->data['Doc']['value'] = 0.01;        
       $this->request->data['Doc']['name'] = 'OneName';
       $this->render('path/to/your/send');
    }        

    if ($this->request->is('post')) {
        $this->set('generated',$this->generate($this->request->data));
        $this->render('path/to/your/generate');
    }       
}

然后在generate.ctp

中

<?php echo $generated ?>

Cakephp将html内容发送到空白视图

1 个答案: