如何使用我的特殊链接替换所有链接? [英] How do I replace all links with my special link?
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问题描述
我正在尝试使用正则表达式替换所有链接甚至相对链接与https://snowycinccino.com/iframe/?url=(这里的旧URL和URL的基础,如果它包括)我需要它与href和src。
我尝试了什么:
这是我目前的代码:
<?php
$ newurl = parse_url($ _ GET [url]);
$ base = $ newurl [scheme]。://。$ newurl [host]。/;
$ basenew = $ newurl [scheme]。://。$ newurl [host];
?>
<?php
$ input = $ page = file_get_contents($ _ GET [url]);
$ domain = $ base;
$ rep ['/ href =(?!https?:\ / \ /)(?! data :)(?!#)/'] ='href ='。$ domain;
$ rep ['/ src =(?!https?:\ / \ /)(?! data :)(?!#)/'] ='src ='。$ domain;
$ rep ['/ @ import [\ n + \ s +]\ //'] ='@ import'。$ domain;
$ rep ['/ @ import [\ n + \ s +]\。/'] ='@ import'。$ domain;
$ output = preg_replace(
array_keys($ rep),
array_values($ rep),
$ input
);
echo $ output;
?>
解决方案
newurl = parse_url(
_GET [url ]);
基=
I am trying to use regex to replace all links even relative links with https://snowycinccino.com/iframe/?url=(Old URL here and base of url if it includes) I need it to work with href and src.
What I have tried:
This is my current code:
<?php $newurl=parse_url($_GET["url"]); $base=$newurl["scheme"]."://".$newurl["host"]."/"; $basenew=$newurl["scheme"]."://".$newurl["host"]; ?> <?php $input = $page = file_get_contents($_GET["url"]); $domain = $base; $rep['/href="(?!https?:\/\/)(?!data:)(?!#)/'] = 'href="'.$domain; $rep['/src="(?!https?:\/\/)(?!data:)(?!#)/'] = 'src="'.$domain; $rep['/@import[\n+\s+]"\//'] = '@import "'.$domain; $rep['/@import[\n+\s+]"\./'] = '@import "'.$domain; $output = preg_replace( array_keys($rep), array_values($rep), $input ); echo $output; ?>
解决方案
newurl=parse_url(
_GET["url"]);
base=
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