Question

我对此很新。我想使用PHP从页面中提取一个表，并在修改所有锚点的HREF值后返回它的HTML。这是表格：

    <html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=windows-1255">
    <link rel="stylesheet" type="text/css" href="../CssGraduateE.css">
    <title></title>
</head>
<body>
    <div>
        <br>
        <table class="main" cellspacing="0" cellpadding="0">
            <tbody>
                <tr>
                    <td>
                        <br><span class="MainHeader">Subjects in Faculty -   Electrical Engineering</span><br><br>
                        <table cellpadding="2" cellspacing="0" border="1" width="100%">
                            <tbody>
                                <tr>
                                    <td><span class="SecondHeader"> Subject Number</span></td>
                                    <td><span class="SecondHeader">Subject   Name</span></td>
                                    <td><span class="SecondHeader">Points</span></td>
                                    <td><span class="SecondHeader">Semesters</span></td>
                                    <td>Subject Site</td>
                                </tr>
                                <tr>
                                    <td><a href="../Subjects/?SUB=46001">46001</a>&nbsp;</td>
                                    <td nowrap="">Engineering of Distributed Software Sys</td>
                                    <td>3</td>
                                    <td><br></td>
                                    <td><a target="_newtab" href="http://www.thislinkisok.com/courses/046001">www</a></td>
                                </tr>
                                <tr>
                                    <td><a href="../Subjects/?SUB=46002">46002</a>&nbsp;</td>
                                    <td nowrap="">Design and Analysis of Algorithms</td>
                                    <td>3</td>
                                    <td>B<br></td>
                                    <td>&nbsp;<br></td>
                                </tr>
                            </tbody>
                        </table>
                    </td>
                </tr>
            </tbody>
        </table>
        <br>
        <table border="0">
            <tbody>
                <tr>
                    <td>Last Update on :</td>
                    <td>Wednesday ,9 April 2014</td>
                    <td></td>
                </tr>
            </tbody>
        </table>
    </div>
</body>
</html>

我知道如何抓住我想要的桌子： $ query = $ xpath-＆gt; query（'// table [@ class =“main”] // table [1]'）; 但是如何循环遍历以“../xxx”开头的所有链接并将其修改为这样的内容：“www.mynewlink.com/xxx”？最后，我想将提取的表作为HTML返回。如何使用本机DOMDocument和DOMXpath执行此操作？

全部谢谢！

Answer 1

如果$html是您从外部网站获取HTML的字符串，则可以执行以下操作：

$dom = new DOMDocument();
@$dom->loadHTML($html);

$xpath = new DOMXPath($dom);

foreach($xpath->query('//table[@class="main"]//a[starts-with(@href, "../")]') as $link) {
    $link->setAttribute('href', preg_replace('#^..#', 'http://www.mynewlink.com', $link->getAttribute('href')));
}

$container = new DOMDocument();
$container->appendChild($container->importNode($xpath->query('//table[@class="main"]')->item(0), true));

echo $container->saveHTML();

使用DOMdocument和DOMXpath刮取网页

1 个答案: