从外部网站获取元素内的图像 URL - Laravel

Question

正如我提到的问题here。我将尝试获取图片网址的替代方法。我想从https://www.matchesfashion.com/products/Adidas-By-Stella-McCartney-Metallic-zebra-print-Primegreen-leggings-1424516 获取产品图片网址，如果您检查产品图片，它可以在<figure></figure> 元素内访问。我做了一些研究并编写了这段代码来从外部网页获取内容。但它没有返回任何东西。

$doc = new DOMDocument;


$doc->preserveWhiteSpace = false;


$doc->strictErrorChecking = false;
$doc->recover = true;

$doc->loadHTMLFile('https://www.matchesfashion.com/products/Adidas-By-Stella-McCartney-Metallic-zebra-print-Primegreen-leggings-1424516');

$xpath = new DOMXPath($doc);

$var = $xpath->evaluate('string(//figure[@class="iiz"])');

我只需要获取该图像的源 URL，这样我就可以继续我的图像编码过程。提前致谢

Answer 1

您好，您可以使用下面的代码来获取图片网址

 $doc = new DOMDocument;
        $doc->preserveWhiteSpace = false;
        $doc->strictErrorChecking = false;
        $doc->recover = true;

        ini_set('user_agent', 'My-Application/2.5');
        libxml_use_internal_errors(true);
        $doc->loadHTMLFile('https://www.matchesfashion.com/products/Adidas-By-Stella-McCartney-Metallic-zebra-print-Primegreen-leggings-1424516');
        $xpath = new DOMXPath($doc);
        $imgs  = $xpath->query('//*[@class="iiz__img "]');
        foreach($imgs as $img)
        {
            echo 'ImgSrc: https:' . $img->getAttribute('src') .'<br />' . PHP_EOL;
        }

这是您想要的结果

ImgSrc：https://assetsprx.matchesfashion.com/img/product/920/1424516_1.jpg

ImgSrc：https://assetsprx.matchesfashion.com/img/product/920/1424516_1.jpg