我有一個解析 img 和文本的代碼。運行php文件中的代碼。它只是顯示 img src、abc、img src、dfe。而且我的代碼不規(guī)則。img 標簽可能帶有鏈接。我想解析 img 和下一個html。像這樣：Array( [0] => Array ( [src] => http://www.whatever.com [text] => abc <br> <h3>title</h3> <div class="content">content <a href="link">my link</a></div> ) [1] => Array ( [src] => http://goingnowhere.com [text] => def <br> <h3>title 2</h3> <div class="content">content <a href="link">my link</a> bla bla bla </div> ))我怎樣才能做到這一點？我目前的代碼：<?php $sample_html = '<img src="http://www.whatever.com" alt="" />abc<br><h3>title</h3><div class="content">content <a href="link">my link</a></div><img src="http://goingnowhere.com" alt="">def<br><h3>title 2</h3><div class="content">content <a href="link">my link</a>bla bla bla</div>';$dom = new DOMDocument();$dom->loadHTML($sample_html);$data = array();$images = $dom->getElementsByTagName('img');foreach ($images as $image) {$data[] = array('src' => $image->getAttribute('src'),'text' => trim($image->nextSibling->textContent),);}echo '<pre>';print_r($data); ?>
1 回答
嗶嗶one
TA貢獻1854條經(jīng)驗 獲得超8個贊
使用 xpath 遍歷所有節(jié)點并使用兩個 img 標簽檢索數(shù)據(jù)。
<?php $sample_html = '
<img src="http://www.whatever.com" alt="" />
abc
<br>
<h3>title</h3>
<div class="content">content <a href="link">my link</a></div>
<img src="http://goingnowhere.com" alt="">
def
<br>
<h3>title 2</h3>
<div class="content">content <a href="link">my link</a>
bla bla bla
</div>
';
$dom = new DOMDocument();
@$dom->loadHtml($sample_html);
$xpath = new DOMXPath($dom);
$snippet = '';
$arr = array();
$count = $xpath->query('//img')->length;
//loop through all img tags
for($i=0;$i<$count;$i++){
$node = $xpath->query('//img')->item($i);
$img_src = $node->getAttribute('src');//first image src
while ($node = $node->nextSibling) {
if (get_class($node) != 'DOMElement') {
continue;
}
if ($node->tagName == 'img') {
$snippet .= $dom->saveXML($node);
$arr[] = array(
'src'=>$img_src,
'content'=>$snippet
);
$img_src = $node->getAttribute('src');//last img src
$snippet = '';
break;
}
$snippet .= $dom->saveXML($node);
}
}
//fill last img data
$arr[] = array('src'=>$img_src,'content'=>$snippet);
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