Symfony提交到相同的URL [英] Symfony submit to same url

问题描述

我有一个带有一些文本字段的表单，并且有一个预览按钮，需要将表单提交给同一控制器.然后在控制器中，我需要提取这些值，并使用这些值填充表单以供模板查看.实现此目标的最佳方法是什么?我是新手，请注意.

I have a form with some text fields and I have a preview button that needs to submit the form to the same controller. And then in the controller, I need to extract the values and populate a form with these values for the template to see. What is the best way to achieve this? I'm a newbe so please be clear.

推荐答案

示例控制器:

public function myControllerName(sfWebRequest $request)
{
  $this->form = new myFormClass();
}

在模板中使用<?php echo $form->renderFormTag( url_for('@yourRoutingName'), array('method' => 'POST') ); ?>，然后将@yourRoutingName更改为指向控制器的那个.

Use <?php echo $form->renderFormTag( url_for('@yourRoutingName'), array('method' => 'POST') ); ?> in your template and change @yourRoutingName to the one pointing to your controller.

现在将您的控制器更改为以下内容:

Now change your controller to be something like this:

public function myControllerName(sfWebRequest $request)
{
  $this->form = new myFormClass();

  if ($request->isMethod(sfRequest::POST)
  {
    $this->form->bind( $request->getParameter( $this->form->getName() ) );

    // Check if the form is valid.
    if ($this->form->isValid())
    {
      $this->form->save();
      // More logic here.
    }
  }
}

$this->form->bind( $request->getParameter( $this->form->getName() ) );部分将发布的数据绑定到您的表单，其中$this->form->isValid()返回一个布尔值，表明该表单是否有效.

The $this->form->bind( $request->getParameter( $this->form->getName() ) ); part binds posted data to your form where $this->form->isValid() returns a boolean whether the form is valid or not.

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