使用简单html dom解析图像src

Question

我正在尝试从此链接获取src图像：https://www.scribd.com/book/348571030/The-Alice-Network-A-Novel

这是什么，但我没有运气

<?php

 include('simple_html_dom.php');

 $html = file_get_html('https://www.scribd.com/book/348571030/The-Alice-Network-A-Novel');

 $list = $html->find('div[class="class="auto__base_component auto__shared_react_document_image react_document_image""]',0);

 $list_array = $list->find('img');

 $list_array2['thumb']  = $list_array->find('img.loaded', 0)->src;

 for ( $i = 0; $i < sizeof($list_array2); $i++ ){
  echo $list_array2[$i]->plaintext;
  echo "<br>";
 }

?>

Answer 1

实际上，我可以使用php类DOMDocument完成它：

$html = file_get_contents('https://www.scribd.com/book/348571030/The-Alice-Network-A-Novel');
$doc = new DOMDocument();
$doc->loadHTML($html);
$xpath = new DOMXPath($doc);
$src = $xpath->evaluate("string(//img/@src)");
    echo $src;

或获取该页面中的所有图像，然后您可以选择所需的内容：

$html = file_get_contents('https://www.scribd.com/book/348571030/The-Alice-Network-A-Novel');
$doc = new DOMDocument();
$doc->loadHTML($html);
$images = $doc->getElementsByTagName('img');
foreach ($images as $image) {
    echo $image->getAttribute('src') . "\n";
}