忽略preg_replace中的img标签 [英] Ignore img tags in preg_replace

问题描述

我想替换HTML字符串中的一个单词，但是如果该单词在'img'元素的属性中，我想排除该替换.

I want to replace a word in an HTML string, but I want to exclude the replacement if the word was in the attributes of the 'img' element.

示例:

$word = 'google';
$html = 'I like google and here is its logo <img src="images/google.png" alt="Image of google logo" />';

$replacement = '<a href="http://google.com">Google</a>';
$result =  preg_replace('/\s'.($word).'/u', $replacement, $html);

preg_replace还将替换'src'和'alt'属性内的"google"词，我希望它仅替换'img'元素外的词.

preg_replace will also replace the "google" words inside 'src' and 'alt' attributes, I want it to just replace the word outside the 'img' element.

推荐答案

您可以使用丢弃模式.例如，您可以使用如下正则表达式:

You can use the discard pattern. For instance you can use a regex like this:

<.*?google.*?\/>(*SKIP)(*FAIL)|google

工作演示

Working demo

此模式背后的想法是丢弃<和>中的google单词，但保留其余部分:

The idea behind this pattern is to discard the google word inside < and > but keep the rest:

<.*?google.*?\/>(*SKIP)(*FAIL)  --> This part will skip the matches where google is within <...>
|google                         --> but will keep the others google

您可以添加所需的许多丢弃"模式，例如:

You can add many "discard" pattern you want, like:

discard patt1(*SKIP)(*FAIL)|discard patt(*SKIP)(*FAIL)|...(*SKIP)(*FAIL)|keep this

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