如何在Zend Framework 2中实现应用程序/网站范围的表单?

时间:2013-04-10 11:29:05

标签: view zend-framework2 zend-form zend-form2

在我的ZF2应用程序中,我有多个模块,其中一个是“搜索”。它提供了搜索表单,搜索逻辑和数据库中的课程列表:

表单类

namespace Search\Form;

...

class CourseSearchForm extends Form {

    private $cities;

    public function __construct(array $cities) {
        parent::__construct('courseSearch');
        $this->setCities($cities);
        $this->setAttribute('method', 'post');
        $this->add(array(
            'name' => 'keyword',
            'attributes' => array(
                'type'  => 'text',
            ),
        ));
        $this->add(array(
            'name' => 'city',
            'type'  => 'Zend\Form\Element\Select',
            'options' => array(
                'label' => 'Stadt',
                'value_options' => $this->cities
            ),
        ));
        // ...
    }

    public function setCities(array $cities) {
        $this->cities = $cities;
    }

}

控制器类

namespace Search\Controller;

...

class SearchController extends AbstractActionController {

    protected $courseTable;

    public function searchAction() {
        return $this->redirect()->toRoute('search-courses');
    }

    public function searchCoursesAction() {
        $form = $this->getServiceLocator()->get('Search\Form\CourseSearchForm');
        $request = $this->getRequest();
        if ($request->isPost()) {
            $courseSearchInput = new CourseSearchInput();
            $form->setInputFilter($courseSearchInput->getInputFilter());
            $form->setData($request->getPost());
            if ($form->isValid()) {
                $courseSearchInput->exchangeArray($form->getData());
                $courses = $this->getCourseTable()->findAllByCriteria($courseSearchInput);
            } else {
                $courses = null;
            }
        } else {
            $courses = null;
        }
        return new ViewModel(array(
            'form' => $form,
            'courses' => $courses,
            'cities' => $this->getServiceLocator()->get('Cache\Model\CityStorage')->getCities(),
        ));
    }

    /**
     * Gets a CourseTable object.
     * @return Search\Model\CourseTable
     */
    function getCourseTable() {
        if (!$this->courseTable) {
            $serviceLocator = $this->getServiceLocator();
            $this->courseTable = $serviceLocator->get('Search\Model\CourseTable');
        }
        return $this->courseTable;
    }

}

/module/Search/view/search/search/search-courses.phtml

中的列表视图
...
<?php
$form = $this->form;
$form->setAttribute('action', $this->url('search-courses'));
$form->prepare();

echo $this->form()->openTag($form);
echo $this->formRow($form->get('keyword'));
echo $this->formRow($form->get('city'));
// ...
echo $this->formSubmit($form->get('submit'));
echo $this->form()->closeTag();
?>
<?php
if ($courses instanceof Zend\Db\ResultSet\ResultSetInterface) {
    echo $this->render('search/search/list-courses.phtml', array(
        'courses' => $courses
    ));
}
?>

现在,搜索表单应显示在网站的每个页面上。如何实现这样的全局表单视图?是否有“标准”方式/最佳做法?

THX

2 个答案:

答案 0 :(得分:0)

您只需创建ViewHelper即可。您将Form注入ViewHelper,然后渲染View并返回它。 ViewHelper本身看起来与ZfcUser在ZfcUser\View\Helper\ZfcUserLoginWidget

中提供的非常相似

答案 1 :(得分:0)

显示它的某种方式:

1。创建自己的视图助手

激励你,

2。使用插件

    // menu user
    $menu = $this->forward()->dispatch('edit_controller', array(
        'action'    => 'menu',
        'item' => 'languages',
        'id' => $id,
    ));
    $viewmodel->addChild($menu, 'menu');

我更喜欢使用plugin forward