使用Zend Framework 2.1的RESTful Web服务.* - 未显示JsonModel

Question

我阅读了几个教程,用ZF2创建一个安静的Web服务.我看到最后一次改变了ZF2如何处理版本2.0.4中发生的restful webservices.让我开始的最有希望的文章是:http: //dustint.com/post/543/getting-started-with-abstractrestfulcontroller

无论如何,我无法完成它,在我看来,在RestController.getList()中,我返回的JsonModel不能像预期的那样工作.由于我的调试调用,我可以识别我的RestController.getList() - 方法将被调用.所有相关代码都在我的github-repository中:https: //github.com/Jochen1980/EhcServer/blob/master/module/Application/src/Application/Controller/RestController.php

class RestController extends AbstractRestfulController{
    public function indexAction(){
        Debug::dump("indexAction()");
        return new ViewModel();
    }
    public function getList() {
        Debug::dump("getList()");
        return new JsonModel(array(
            array('name' => 'test'),
            array('name' => 'second')
        ));
    } 
    ...

目前我收到此错误消息:致命错误:未捕获异常'Zend\View\Exception\RuntimeException',消息'Zend\View\Renderer\PhpRenderer :: render:无法呈现模板"application/rest/get-list"; 解析器无法解析为第499行的C:\ xampp\htdocs\EhcServer\vendor\zendframework\zendframework\library\Zend\View\Renderer\PhpRenderer.php中的文件

提前致谢!

Answer 1

你strategies需要在里面view_manager的module.config.php

也就是说,视图管理器部分应该如下所示

'view_manager' => array(
    'display_not_found_reason' => true,
    'display_exceptions' => true,
    'doctype' => 'HTML5',
    'not_found_template' => 'error/404',
    'exception_template' => 'error/index',
    'template_map' => array(
        'layout/layout' => __DIR__ . '/../view/layout/layout.phtml',
        'application/index/index' => __DIR__ . '/../view/application/index/index.phtml',
        'error/404' => __DIR__ . '/../view/error/404.phtml',
        'error/index' => __DIR__ . '/../view/error/index.phtml',
    ),
    'template_path_stack' => array(
        __DIR__ . '/../view',
    ),
    // let the view manager know which strategies to use
    'strategies' => array(
        'ViewJsonStrategy',
    ),
),