Question

您好我有一个Zend Framework应用程序，以下代码是popularAction控制器

public function popularAction()
    {
        $type = $this->_getParam('ref',1);
        if($type == 'reviews'){
        $businessReviewMapper = new Application_Model_Mapper_BusinessReviewsMapper();
        $result = $businessReviewMapper->getTotalVote();

        $rotd = $businessReviewMapper->getROTD($result['review_id']);
        $rotd[0]['u_img'] = $this->view->getLoginUserImage($rotd[0]['social_id'],$rotd[0]['login_type'],null,null,square);
        $rotd[0]['rating'] = $this->view->getRatingImg($rotd[0]['rating']);
        $rotd[0]['business_name_url'] = preg_replace("![^a-z0-9]+!i","-", $rotd[0]['business_name']);            
        $this->render('reviews');
        $this->_helper->json($rotd);



        } elseif($type == 'openings') {
            $this->view->text = "New Openings";

        } else {
            $this->_helper->redirector('index', 'index', 'default');
        }

    }

当用户浏览到http://localhost/business/popular?ref=reviews时，上述控制器代码将呈现reviews.phtml模板。现在在模板本身内部存在对数据的ajax请求，如下所示：

function getPopular()
{
    var count=1;
    $.ajax({
        url:"<?=$this->baseUrl('business/popular?ref=reviews')?>",
        data:{'count':count},
        dataType:"json"
        type:"POST",
        success:function(data){
            alert('ok')
        }
    });

遗憾的是$this->_helper->json($rotd);没有将数据传递给reviews.phtml，但显示了zend db model返回的json数据，我可能错了？谢谢

Answer 1

如果您的目标是，要在Ajax请求上发送JSON而不是.phtml文件，请尝试实现此目的：

  if ($this->getRequest()->isXmlHttpRequest()) {
        if ($this->getRequest()-isPost()) {
            $this->_helper->json($rotd);
        }
    }

这部分检查请求是否，让我们调用它：ajax based ..

Zend Json将数据呈现到视图中

1 个答案: