Question

我正在尝试将变量传递给视图，这个非常奇怪，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pr变量上执行outputs，但这会输出到NULL，但是当我删除控制器函数内的整个代码串并只传递一个测试变量时通过它工作。

我不知道该功能有什么问题吗？

顺便说一句，上述功能没有发现任何错误

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变量之前从控制器中的set函数设置之前得到的

Answer 1

您选择使用CakePHP的原因是什么？因为您似乎不使用其功能！

您正在使用文字SQL查询，因此基本上不会跳过模型functionality。

您是从Controller输出内容了吗？使用输出缓冲时要小心，这可能与CakePHP的内部工作冲突，后者在很多情况下也依赖于输出缓冲。由于您已在此处输出内容（ob_flush（）），因此您将在到达视图之前输出内容。

通常情况下，我会指出手册中的具体points，因为这里有太多错误，我建议开始阅读beginning

Cakephp设置函数不将变量从控制器传递给视图

1 个答案: