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php解析字符串里所有URL地址的方法

程序员文章站 2024-01-25 14:56:58
本文实例讲述了php解析字符串里所有url地址的方法。分享给大家供大家参考。具体如下:

本文实例讲述了php解析字符串里所有url地址的方法。分享给大家供大家参考。具体如下:

<?php
// $html = the html on the page
// $current_url = the full url that the html came from
//(only needed for $repath)
// $repath = converts ../ and / and // urls to full valid urls
function pagelinks($html, $current_url = "", $repath = false){
  preg_match_all("/\<a.+?href=(\"|')(?!javascript:|#)(.+?)(\"|')/i", $html, $matches);
  $links = array();
  if(isset($matches[2])){
    $links = $matches[2];
  }
  if($repath && count($links) > 0 && strlen($current_url) > 0){
    $pathi   = pathinfo($current_url);
    $dir    = $pathi["dirname"];
    $base    = parse_url($current_url);
    $split_path = explode("/", $dir);
    $url    = "";
    foreach($links as $k => $link){
      if(preg_match("/^\.\./", $link)){
        $total = substr_count($link, "../");
        for($i = 0; $i < $total; $i++){
          array_pop($split_path);
        }
        $url = implode("/", $split_path) . "/" . str_replace("../", "", $link);
      }elseif(preg_match("/^\/\//", $link)){
        $url = $base["scheme"] . ":" . $link;
      }elseif(preg_match("/^\/|^.\//", $link)){
        $url = $base["scheme"] . "://" . $base["host"] . $link;
      }elseif(preg_match("/^[a-za-z0-9]/", $link)){
        if(preg_match("/^http/", $link)){
          $url = $link;
        }else{
          $url    = $dir . "/" . $link;
        }
      }
      $links[$k] = $url;
    }
  }
  return $links;
}
header("content-type: text/plain");
$url = "//www.jb51.net";
$html = file_get_contents($url);
// gets links from the page:
print_r(pagelinks($html));
// gets links from the page and formats them to a full valid url:
print_r(pagelinks($html, $url, true));

希望本文所述对大家的php程序设计有所帮助。