如何重新填充此 zend 表单？

Question

我有一个问题，我似乎无法获得重新填充 zend 表单的值。我使用的表格与我习惯的有点不同。

我已将“社会保险号”分成三个字段，值以数组形式出现，但我无法让它们在操作后返回。

这是我的表单类：

class Application_Form_Checkssnforexistingreferral extends Zend_Form {
    public function init() {

        // SSN
        $this->addElement('text', 'ss_number', array(
                'label' => 'SSN: ',
                'required' => True,
        ));

    }
}

然后在我看来，我使用这样的形式......

<?php // SSN (first segment)
                            echo '<p>Social Security Number</p>';
                            echo $this->formText('ss_number["first"]', '', array(
                                'size' => 3, 
                                'maxlength' => 3,
                                'required' => True,
                                'id' => 'ssn1',
                                )) 
                            ?>
                        <span>-</span>
                        <?php // SSN (second segment) 
                            echo $this->formText('ss_number["second"]', '', array(
                                'size' => 2, 
                                'maxlength' => 2,
                                'required' => True,
                                'id' => 'ssn2',
                                )) 
                            ?>
                        <span>-</span>
                        <?php // SSN (third segment) 
                            echo $this->formText('ss_number["third"]', '', array(
                                'size' => 4, 
                                'maxlength' => 4,
                                'required' => True,
                                'id' => 'ssn3',
                                )) 
                            ?>

我这样做是为了更好地控制表单元素的样式和呈现方式，它在较大的表单上运行良好，尽管我在填充该表单上的字段时也遇到了问题。

这是我在控制器中尝试的...

if ($this->getRequest()->isPost()) {
            $formData = $this->getRequest()->getPost();
            if ($form->isValid($formData)) {



                $referralsModel = new Application_Service_Findssninreferrals();
                $referrals = $referralsModel->findSocialSecurityNumber($formData);

                // load the view's parameter 'referral' w/ the object collection
                // and 'NULL' the 'first page load' parameter
                $this->view->referrals = $referrals;

                $first = $formData['ss_number']['"first"'];
                $second = $formData['ss_number']['"second"'];
                $third = $formData['ss_number']['"third"'];

                $form->populate(array('ss_number["first"]' => $first,
                                      'ss_number["second"]' => $second,
                                      'ss_number["third"]' => $third
                            ));
                if (empty($referrals)) {
                    $flashMessenger->addMessage('There is no record found for this SSN, you may create a new referral for this client');

                    print_r($formData);
                    $form->populate(array($formData['ss_number']));

                    $ssn = $first . $second . $third;
                    $this->view->continueLink = "link to create new referral" . $ssn;
                    } 

            } else {
                      // else populate the form and allow the correction of...
                      $flashMessenger->addMessage('There was a problem with the number that you entered, please try again...');
                      $form->populate($formData);
                   }

            }
            $this->view->form = $form;

...

这就是其中一个元素呈现为 HTML 的方式...

<p>Social Security Number</p>
<input type="text" required="1" maxlength="3" size="3" value="" id="ssn1" name="first" class="idleField">

在控制器中，'if(emtpy($referrals))' 部分是我进行大部分实验的地方，试图让它重新填充字段。上面的部分也不起作用，我基本上试图只是一个 'form->populate(array(...' 但也没有运气。我只是没有从 'populate' 方法中得到任何东西......

Answer 1

尝试重定向到请求的 url 应该会立即返回并填充表单。否则这可能行不通，你不能在没有请求的情况下填充表单（可以使用 ajax），并且在发布后你没有新的请求。这一个动作可能至少应该是 2 个动作，然而 你可以这样试试：

 if ($this->getRequest()->isPost()) {
            $formData = $this->getRequest()->getPost();
            if ($form->isValid($formData)) {

                $referralsModel = new Application_Service_Findssninreferrals();
                $referrals = $referralsModel->findSocialSecurityNumber($formData);

                // load the view's parameter 'referral' w/ the object collection
                // and 'NULL' the 'first page load' parameter
                $this->view->referrals = $referrals;

                $first = $formData['ss_number']['"first"'];
                $second = $formData['ss_number']['"second"'];
                $third = $formData['ss_number']['"third"'];

                if (empty($referrals)) {
                    $flashMessenger->addMessage('There is no record found for this 
                        SSN, you may create a new referral for this client');

                    $this->_redirect($this->getRequest()->getRequestUri());
                    //next line may or may not be needed or help
                    $form->populate($formData);

                    $ssn = $first . $second . $third;
                    $this->view->continueLink = "link to create new referral" . $ssn;
                }
            } else {
                // else populate the form and allow the correction of...
                $flashMessenger->addMessage('There was a problem with the number that you entered, please try again...');
                $form->populate($formData);
            }
        }
        $this->view->form = $form;