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我有一个对我来说似乎很难的非常重要的问题.. 只是我需要从当前节点的父节点中选择子节点!我知道我们使用 (..) 来选择父节点。但就我而言,这还不够..这是我获取所有颜色值的 php 代码

<?php
    /////////////////////////////////////////////////////////////////////
        $html='
            <table>
                <tr>
                    <td colspan="2">
                        <span>green</span>
                        <img src="green.gif" />
                    </td>
                </tr>
                <tr>
                    <td>
                        <span>yellow</span>
                        <img src="yellow.gif" />
                    </td>
                    <td>
                        <span>red</span>
                        <img src="red.gif" />
                    </td>
                </tr>
            </table>
            <table>
                <tr>
                    <td>
                        <span>black</span>
                        <img src="black.gif" />
                    </td>
                </tr>
            </table>
        ';
    /////////////////////////////////////////////////////////////////////
        $dom = new DOMDocument();
        $dom->loadHTML($html);
        $xpath = new DomXPath($dom);
    /////////////////////////////////////////////////////////////////////
        $evaluate = $xpath->evaluate('.//table/tr/td/span');
        for($x=0,$results=''; $x<$evaluate->length; $x++)
        {
            $x1=$x+1;

            $query1 = $xpath->query('.//table/tr/td/span');
            $color = $query1->item($x)->nodeValue;
            $results .= "color $x1 is : $color<br/>";
        }
        echo $results;
    /////////////////////////////////////////////////////////////////////
?>

* now i need to get for every color its images src ..
i tried this but no way 

<?php
    /////////////////////////////////////////////////////////////////////
        $html='
            <table>
                <tr>
                    <td colspan="2">
                        <span>green</span>
                        <img src="green.gif" />
                    </td>
                </tr>
                <tr>
                    <td>
                        <span>yellow</span>
                        <img src="yellow.gif" />
                    </td>
                    <td>
                        <span>red</span>
                        <img src="red.gif" />
                    </td>
                </tr>
            </table>
            <table>
                <tr>
                    <td>
                        <span>black</span>
                        <img src="black.gif" />
                    </td>
                </tr>
            </table>
        ';
    /////////////////////////////////////////////////////////////////////
        $dom = new DOMDocument();
        $dom->loadHTML($html);
        $xpath = new DomXPath($dom);
    /////////////////////////////////////////////////////////////////////
        $evaluate = $xpath->evaluate('.//table/tr/td/span');
        for($x=0,$results=''; $x<$evaluate->length; $x++)
        {
            $x1=$x+1;

            $query1 = $xpath->query('.//table/tr/td/span');
            $color = $query1->item($x)->nodeValue;

            $query2 = $xpath->query('.//table/tr/td/span['.$x1.']/../img');
            $image = $query2->item($x)->getAttribute("src");
            $results .= "color $x1 is : $color - and- image $x1 is : $image<br/>";
        }
        echo $results;
    /////////////////////////////////////////////////////////////////////
?>

请帮忙 ::

4

1 回答 1

4

试试这个:

... initiate/load dom+xpath ...
$res = $xpath->query('//table/tr/td');
foreach($res as $td) {
    $spans = $td->getElementsByTagName('span');
    $color = $spans[0]->nodeValue;
    $imgs = $td->getELementsByTagName('img');
    $src = $imgs[0]->getAttribute('src');
}

无需两次运行完整的 xpath 查询。DOM 树中的每个节点都完全了解其父节点和子节点。

于 2011-08-14T05:28:41.773 回答