Question

我有一个带有test.htm名称的html文件，其代码如下：

<div></div>

<div>
<div class="body">
<div>
<a href="login.php"><img src="./lklk_files/81755_236.jpg"></a>
</div> 

<div class="wrapper"></div> 

</div>

<div class="photo-holder"></div>
</div>

我想仅从具有“body”属性值的class属性的DIV中提取img src值我使用下面的PHP代码：

<?php

$f = "test.htm";
$doc = new DOMDocument();
$doc->loadHTMLFile($f);
$xpath = new DOMXPath($doc);

$nodeList = $xpath->query("//div");
foreach($nodeList as $node)
if($node->getAttribute('class') == "body"){
$s = $node->nodeValue;
$doc2 = new DOMDocument();
$doc2->loadHTML($s);
$xpath2 = new DOMXPath($doc2);
$img = $xpath2->query("//img");
foreach($img as $g)
echo $g->attributes->getNamedItem('src')->nodeValue;
}
?>

但是当它运行时，说你的$ s是一个空字符串。我的代码在哪里出错了？

Answer 1

你能试试这个，$domxpath->evaluate("string(//img/@src)");

    $htmlString= '<div></div>

    <div>
    <div class="body">
    <div>
    <a href="login.php"><img src="./lklk_files/81755_236.jpg"></a>
    </div> 

    <div class="wrapper"></div> 

    </div>

    <div class="photo-holder"></div>
    </div>';

    $dom = new DOMDocument();
    $dom->loadHTML($htmlString);
    $domxpath = new DOMXPath($dom);
    echo $src = $domxpath->evaluate("string(//img/@src)");

Answer 2

这样做......

<?php
$html='<div></div>

<div>
<div class="body">
<div>
<a href="login.php"><img src="./lklk_files/81755_236.jpg"></a>
</div>

<div class="wrapper"></div>

</div>

<div class="photo-holder"></div>
</div>';
    $dom = new DOMDocument;
    $dom->loadHTML($html);
foreach ($dom->getElementsByTagName('div') as $tag) {

    if($tag->getAttribute('class')=='body')
    {
         foreach($tag->getElementsByTagName('img') as $imgtag)
        {
            echo $imgtag->getAttribute('src');
            exit;
        }
    }
}

输出：

./lklk_files/81755_236.jpg

使用php从html文件中的特殊div中提取图像src值

2 个答案: