Cakephp 设置函数不将变量从控制器传递到视图

Question

我正在尝试将变量传递给视图，这个变量非常奇怪，因为命名和目录结构是正确的。以下是我的控制器中的功能：

public function validate_apply_link(){
        App::uses('CakeEmail', 'Network/Email');

        $this->layout = 'blank';

        $listings = $this->CareersAndJob->query("
            SELECT l.sid, l.title, lp.value, u.CompanyName, u.WebSite
            FROM listings l
            LEFT JOIN listings_properties lp
            ON lp.object_sid = l.sid
            LEFT JOIN users u
            ON u.sid = l.user_sid
            WHERE l.active = 1
            AND lp.add_parameter = 2
            AND l.JobGateSenderReference IS NULL
            AND u.CompanyName != 'AECOM'
            ORDER BY u.CompanyName ASC
            LIMIT 5
            ");

        $doc = new DOMDocument();
        ob_start();
        $listing_count = count($listings);
        echo nl2br("Checking $listing_count active jobs...\n\n");
        $i=0;
        foreach($listings as $listing){

            $sid = $listing['l']['sid'];
            $url = $listing['lp']['value'];
            $company_name = $listing['u']['CompanyName'];
            $title = htmlspecialchars($listing['l']['title']);

            $length = strpos($title, "-");
            if($length != 0){
                $title = substr($title, 0, $length-1);
            }
            $title = substr($title, 0, $length-1);
            $title = substr($title, 0, 10);
            $data = $this->curl($url);

            $check_pdf = strpos($data['info']['content_type'], "pdf");
            if($check_pdf != false){
                $outputs['data'][$i]['url'] = $url;
                $outputs['data'][$i]['sid'] = $sid;
                $outputs['data'][$i]['title'] = $title;
                $outputs['data'][$i]['company_name'] = $company_name;
                $outputs['data'][$i]['our_link'] = "http://careersandjobs.com.au/display-job/{$sid}";
                $outputs['data'][$i]['content_type'] = $data['info']['content_type'];

                $outputs['data'][$i]['data_type'] = 'pdf';

                $i++;
                continue;
            }

            @$doc->loadHTML($data['results']);
            $html = $doc->saveHTML();
            $xpath = new DOMXpath($doc);

            $body = $doc->getElementsByTagName('body')->item(0);

            $parsed_url = parse_url($url);

            switch($parsed_url['host']){
                case "www.michaelpage.com.au":
                    parse_str($url);
                    $exist = $xpath->query("//*[contains(@value,'{$ref}')]");

                break;
                case "https://vacancies.mackay.qld.gov.au":
                    parse_str($url);
                    $exist = $xpath->query("//*[contains(@value,'{$title}')]");
                break;
                default:
                    $exist = $xpath->query("//*[contains(text(),'{$title}')]");
                break;
            }

            if($exist->length == 0){

                if(strpos($url, '#') == false){

                $outputs['data'][$i]['url'] = $url;
                $outputs['data'][$i]['sid'] = $sid;
                $outputs['data'][$i]['title'] = $title;
                $outputs['data'][$i]['company_name'] = $company_name;
                $outputs['data'][$i]['our_link'] = "http://careersandjobs.com.au/display-job/{$sid}";
                $outputs['data'][$i]['content_type'] = $data['info']['content_type'];

                $response_code = $this->http_response_codes($data['info']['http_code']);
                $outputs['data'][$i]['response_code'] = $response_code;

                $outputs['data'][$i]['data_type'] = 'title_not_found';


                }else{

                    $outputs['data'][$i]['data_type'] = 'no_iframe';
                }

                $i++;
            }

            flush();
            ob_flush();

        }

        $this->set(compact('outputs'));

    }

我可以在视图中的outputs 变量上执行pr，但这会输出到NULL，但是当我删除控制器函数内的整组代码并通过它传递一个测试变量时，它就可以工作了。

我不知道的功能有什么问题吗？

顺便说一下上面的函数没有发现错误

app/Controller/CareersAndJobsController.php (line 1048)
array(
    'data' => array(
        (int) 0 => array(
            'url' => 'http://bawbawshire.currentjobs.com.au/cvbuilder/apply+for+this+job/no/1225055',
            'sid' => '3649',
            'title' => 'Graduate P',
            'company_name' => 'Baw Baw Shire Council',
            'our_link' => 'http://careersandjobs.com.au/display-job/3649',
            'content_type' => 'text/html; charset=utf-8',
            'response_code' => 'OK',
            'data_type' => 'title_not_found'
        ),
        (int) 1 => array(
            'url' => 'http://bawbawshire.currentjobs.com.au/cvbuilder/apply+for+this+job/no/1225724',
            'sid' => '3726',
            'title' => 'Program &a',
            'company_name' => 'Baw Baw Shire Council',
            'our_link' => 'http://careersandjobs.com.au/display-job/3726',
            'content_type' => 'text/html; charset=utf-8',
            'response_code' => 'OK',
            'data_type' => 'title_not_found'
        ),
        (int) 2 => array(
            'url' => 'http://bawbawshire.currentjobs.com.au/cvbuilder/apply+for+this+job/no/1225826',
            'sid' => '3727',
            'title' => 'Road Netwo',
            'company_name' => 'Baw Baw Shire Council',
            'our_link' => 'http://careersandjobs.com.au/display-job/3727',
            'content_type' => 'text/html; charset=utf-8',
            'response_code' => 'OK',
            'data_type' => 'title_not_found'
        )
    )
)

这是我在 outputs 变量被控制器中的 set 函数设置之前得到的结果

Answer 1

您选择使用 CakePHP 的任何原因？因为您似乎没有使用它的功能！

您使用的是文字 SQL 查询，因此基本上跳过了模型 functionality。

您是从Controller 输出内容吗？使用输出缓冲时要小心，这可能与 CakePHP 的内部工作相冲突，CakePHP 在许多情况下也依赖于输出缓冲。因为您已经在此处输出内容（ob_flush()），所以您将在到达您的视图之前输出您的内容..

通常我会在手册中指出具体的points，但是，因为这里有很多错误，我建议从beginning开始阅读