Question

实际上我想在PHP中通过图像alt从HTML代码中获取图像URL [src]，但我没有选择。

<?php
$html = '
<img border="0" src="/images/image11.jpg" alt="Image11" width="100" height="100" />
<img border="0" src="/images/image22.jpg" alt="Image22" width="100" height="100" />';

$doc = new DOMDocument();
$doc->loadHTML($html);
$xpath = new DOMXPath($doc);
$src = $xpath->evaluate("string(//img/@src)");

// will return /images/image.jpg
echo $src;
?>

此代码只是获得图像的src，其中包含alt =＆＃34; Image22＆＃34;。

Answer 1

$html = '<img border="0" src="/images/image11.jpg" alt="Image11" width="100" height="100" /><img border="0" src="/images/image22.jpg" alt="Image22" width="100" height="100" />';

$dom = new DomDocument();
$dom->loadHTML($html);

$images = $dom->getElementsByTagName('img');

foreach ($images as $key => $image){

    if ($image->hasAttributes()){

        if ($image->hasAttribute("src")){

            if ($image->hasAttribute("alt")){

                $alt = $image->getAttribute("alt");

                if ($alt == 'Image22'){

                    $src = $image->getAttribute("src");

                    echo "<br>" . $src . "<br>";

                }

            }

        }

    }

}

Answer 2

<?php
$html = '
<img border="0" src="/images/image11.jpg" alt="Image11" width="100" height="100" />
<img border="0" src="/images/image22.jpg" alt="Image22" width="100" height="100" />';

$doc = new DOMDocument();
$doc->loadHTML($html);
$xpath = new DOMXPath($doc);

$res = $xpath->query('//img');
foreach($res as $img) {

    echo $img->getAttribute('alt');
}


?>

如何在PHP中通过Alt从HTML代码中获取图像URL

2 个答案: