使用正则表达式从字符串中捕获文本之间的所有链接和链接 - Using RegEx to Capture All Links & In Between Text From A String

Using RegEx to Capture All Links & In Between Text From A String

<Link to: http://www.someurl(.+)> maybe some text here(.*) <Link: www.someotherurl(.+)> maybe even more text(.*)

鉴于这一切都在一行上，我如何匹配或更好地提取所有完整的 url 和文本？对于此示例，我希望提取：

http://www.someurl(.+) . maybe some text here(.*) . www.someotherurl(.+) . maybe even more text(.*)

基本上，<Link.*:.*将启动每个链接捕获，>将结束它。然后，第一次捕获后的所有文本也将被捕获，直到下一次链接捕获出现零次或多次。

我试过：

preg_match_all('/<Link.*?:.*?(https|http|www)(.+?)>(.*?)/', $v1, $m4);

但我需要一种方法来捕获结束>后的文本.问题是在第一个链接之后可能有也可能没有另一个链接(当然也可能没有链接！

$string = "<Link to: http://www.someurl(.+)> maybe some text here(.*) <Link: www.someotherurl(.+)> maybe even more text(.*)";
$string = preg_split('~<link(?: to)?:'s*([^>]+)>~i',$string,-1,PREG_SPLIT_DELIM_CAPTURE|PREG_SPLIT_NO_EMPTY);
echo "<pre>";
print_r($string);

输出：

Array
(
    [0] => http://www.someurl(.+)
    [1] =>  maybe some text here(.*) 
    [2] => www.someotherurl(.+)
    [3] =>  maybe even more text(.*)
)

您可以使用

此模式：

preg_match_all('~<link'b[^:]*:'s*'K(?<link>[^'s>]++)[^>]*>'s*(?<text>[^<]++)~',
               $txt, $matches, PREG_SET_ORDER);
foreach($matches as $match) {
    printf("<br/>link: %s'n<br/>text: %s", $match['link'], $match['text']);
}