Php 用换行符显示网页内容

Php 用换行符显示网页内容,php,regex,newline,Php,Regex,Newline,我解析网页内容并以以下方式在div中显示: $doc = new DOMDocument(); libxml_use_internal_errors(true); $doc->loadHTMLFile($url); libxml_clear_errors(); $xpath = new DOMXPath($doc); foreach($xpath->query("//script") as $script) { $script->parentNode->remov

我解析网页内容并以以下方式在div中显示:

$doc = new DOMDocument();
libxml_use_internal_errors(true);
$doc->loadHTMLFile($url);
libxml_clear_errors();
$xpath = new DOMXPath($doc);
foreach($xpath->query("//script") as $script) {
    $script->parentNode->removeChild($script);
}

$textContent = $doc->textContent; //inherited from DOMNode
$text=escapeshellarg($textContent);

$test = preg_replace("/[^a-zA-Z]+/", " ", html_entity_decode($text));
但这也会将换行符(段落)删除到$test中。
如何在上面包含换行符,使每个段落都可见?

PHP有一个很好的函数。试试这个:

echo nl2br($test);

只需不替换新行,即可包含新行

$test = preg_replace("/[^a-zA-Z\n]+/", " ", html_entity_decode($text));
                               ^^

谢谢,但是它跳过了换行符。这是我的代码审查请