Question

I want to transform the following:

1. From:
<img class="lazyjs bbcodeImage" src="//google.com/blank.gif" data-original="http://google.com/poster.jpg" alt="image" />

1. To:
<img src="http://domain.com/poster.jpg" />

2. From:
<a rel="nofollow" href="/confirm/url/aHR0cDovL2dvb2dsZS5jb20%3D/" class="ajaxLink">

2. To:
<a href="http://google.com">

Basically, I want to use the data-original for <img src. The <a href is first encoded with base64_encode, then with urlencode.

Here's what I've done at until now:

<?php
// 1
$string = '<img class="lazyjs bbcodeImage" src="//google.com/blank.gif" data-original="http://google.com/poster.jpg" alt="image" />';

echo preg_replace('/<img class="lazyjs bbcodeImage" src="\/\/google.com\/blank.gif" data-original="(.*?)" alt="image" \/>/', '<img src="$1" />', $string);

// 2
$string = '<a rel="nofollow" href="/confirm/url/aHR0cDovL2dvb2dsZS5jb20%3D/" class="ajaxLink">';

echo preg_replace('/<a rel="nofollow" href="\/confirm\/url\/(.*?)\/" class="ajaxLink">/', '<a href="$1">', $string);
?>

The problem is that on 2 I don't know how to decode the $1.

Answer 1

可能会有人建议regex，但根据RegEx match open tags except XHTML self-contained tags，这不是一个正确的解决方案。感谢上帝有人PHPquery。通过这种方式，您可以像在jQuery中一样使用选择器来选择这些属性。

Answer 2

没关系，我想我做到了：

<?php
// 1
$string = '<img class="lazyjs bbcodeImage" src="//google.com/blank.gif" data-original="http://google.com/poster.jpg" alt="image" />';

echo preg_replace('/<img class="lazyjs bbcodeImage" src="\/\/google.com\/blank.gif" data-original="(.*?)" alt="image" \/>/', '<img src="$1" />', $string);

// 2
$string = '<a rel="nofollow" href="/confirm/url/aHR0cDovL2dvb2dsZS5jb20%3D/" class="ajaxLink">';

echo preg_replace_callback('/<a rel="nofollow" href="\/confirm\/url\/(.*?)\/" class="ajaxLink">/', function ($match) { return '<a href="' . base64_decode(urldecode($match[1])) . '">'; }, $string);
?>

替换<img/> src并解码<a> href

2 个答案: