谷歌网址上的后期处理 [英] post processing on google urls

问题描述

我使用正则表达式从 Google 搜索结果中抓取了一些网址.它为我提供了以下格式的链接.现在，我只想要方案和主机.任何正则表达式的想法?我尝试了 url_parse，但由于前缀/url?q="，它似乎不起作用.

I have grabbed some urls from Google search results using a regular expression. It has provided me the links in the format given below. Now, I just want the scheme and the host. Any regex ideas? I tried url_parse, but it does not seem to work because of the prefix '/url?q='.

/url?q=http://www.fertile-focus.com/&sa=U&ei=dTTTU7L2A4egugSY44LgAQ&ved=0CCsQFjAEOGQ&usg=AFQjCNEwG9ntbG0ZtqbqjJNSfVTlqQJYmg

/url?q=http://www.genetests.org/&sa=U&ei=dTTTU7L2A4egugSY44LgAQ&ved=0CDgQFjAGOGQ&usg=AFQjCNFiux9o5YIUGP4P8B_oG_J6iD1Y6g

现在只需要

http://www.fertile-focus.com
http://www.genetests.org

推荐答案

Regex 匹配上面提到的以 /url?q= 开头的 URL，

Regex to match the above mentioned URL's which are preceded by /url?q= ,

\/url\?q=\K.*?(?=\/&)

演示

或

www\.[^.]*\.(?:org|com)

演示

您的 PHP 代码将是，

Your PHP code would be,

<?php
$url = <<< 'EOT'
/url?q=http://www.fertile-focus.com/&sa=U&ei=dTTTU7L2A4egugSY44LgAQ&ved=0CCsQFjAEOGQ&usg=AFQjCNEwG9ntbG0ZtqbqjJNSfVTlqQJYmg

/url?q=http://www.genetests.org/&sa=U&ei=dTTTU7L2A4egugSY44LgAQ&ved=0CDgQFjAGOGQ&usg=AFQjCNFiux9o5YIUGP4P8B_oG_J6iD1Y6g
EOT;
$regex =  '~\/url\?q=\K.*?(?=\/&)~';
preg_match_all($regex, $url, $matches);
var_dump($matches);
?>

输出:

array(1) {
  [0]=>
  array(2) {
    [0]=>
    string(28) "http://www.fertile-focus.com"
    [1]=>
    string(24) "http://www.genetests.org"
  }
}

这篇关于谷歌网址上的后期处理的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！