在PHP5中,用于进行此转换的良好preg_replace
表达式是什么:
使用<br />
替换换行符,但仅限<pre>
个阻止
(随意做出简化假设,并忽略极端情况。例如,我们可以 假设标签是一行,而不是病态的东西,如)
输入文字:
<div><pre class='some class'>1
2
3
</pre>
<pre>line 1
line 2
line 3
</pre>
</div>
输出:
<div><pre>1<br />2<br />3<br /></pre>
<pre>line 1<br />line 2<br />line 3<br /></pre>
</div>
(激励上下文:尝试在维基词典语法中修改错误20760的语法高亮度扩展,找到我的PHP技能(我主要做的是python)并不符合要求)。
除了regexen之外,我对其他解决方案持开放态度,但是较小的是首选(例如,构建html解析机制是过度的)。
答案 0 :(得分:6)
这样的东西?
<?php
$content = "<div><pre class='some class'>1
2
3
</pre>
<pre>line 1
line 2
line 3
</pre>
</div>
";
function getInnerHTML($Node)
{
$Body = $Node->ownerDocument->documentElement->firstChild->firstChild;
$Document = new DOMDocument();
$Document->appendChild($Document->importNode($Body,true));
return $Document->saveHTML();
}
$dom = new DOMDocument();
$dom->loadHTML( $content );
$preElements = $dom->getElementsByTagName('pre');
if ( count( $preElements ) ) {
foreach ( $preElements as $pre ) {
$value = preg_replace( '/\n|\r\n/', '<br/>', $pre->nodeValue );
$pre->nodeValue = $value;
}
echo html_entity_decode( getInnerHTML( $dom->documentElement ) );
}
答案 1 :(得分:0)
基于SilentGhost所说的内容(由于某种原因这里没有显示):
<?php
$str = "<div><pre class='some class' >1
2
3
< / pre>
<pre>line 1
line 2
line 3
</pre>
</div>";
$out = "<div><pre class='some class' >1<br />2<br />3<br />< / pre>
<pre>line 1<br />line 2<br />line 3<br /></pre>
</div>";
function protect_newlines($str) {
// \n -> <br />, but only if it's in a pre block
// protects newlines from Parser::doBlockLevels()
/* split on <pre ... /pre>, basically. probably good enough */
$str = " ".$str; // guarantee split will be in even positions
//$parts = preg_split('/(<pre .* pre>)/Umsxu',$str,-1,PREG_SPLIT_DELIM_CAPTURE);
$parts = preg_split("/(< \s* pre .* \/ \s* pre \s* >)/Umsxu",$str,-1,PREG_SPLIT_DELIM_CAPTURE);
foreach ($parts as $idx=>$part) {
if ($idx % 2) {
$parts[$idx] = preg_replace("/\n/", "<br />", $part);
}
}
$str = implode('',$parts);
/* chop off the first space, that we had added */
return substr($str,1);
}
assert(protect_newlines($str) === $out);
?>