需要用模板代码替换img src

Question

我需要将多个数据库条目中的所有/media/* url替换为{media url=URL}

我有以下几项测试：

<img src="/media/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />
<img src="/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />
<img src="/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />
<img src="/media/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />

我试图使用下面的Regex

<img.+?src=\"\/media(.+?)[\"'].*?>/g

最终结果应该如下所示：

<img src="{{media url="/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg}}" alt="NLA for Her Her Cleanse 60ct" />

我计划使用php的preg_replace

对于regex，我应该使用什么来查找和替换url？我想在这里弄清楚：https://regex101.com/r/kF3gI1/1

Answer 1

因此，我继续并假设所有条目将使用双引号来封装图像URL。下面是我用来测试的代码：

$tests = array(
    '<img src="/media/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />',
    '<img src="/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />',
    '<img src="/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />',
    '<img src="/media/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />',
);

foreach ($tests as $str) {
    echo preg_replace("/<img.+?src=\"(.+?)\"/","<img src=\"{{media url=\"$1\"}}\"",$str)."\r\n";
}

以下是研究结果：

<img src="{{media url="/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg"}}" alt="NLA for Her Her Cleanse 60ct" />
<img src="/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />
<img src="/nutritionfacts/NLAHCLEA00600000CP.jpg" alt="NLA for Her Her Cleanse 60ct" />
<img src="{{media url="/wysiwyg/nutritionfacts/NLAHCLEA00600000CP.jpg"}}" alt="NLA for Her Her Cleanse 60ct" />