Question

我需要用几个数据库条目替换所有/ media / * url {{media url = URL}}

我有以下测试：

<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" />

我试图使用以下正则表达式

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

最终结果应该是：

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

我打算使用php的preg_replace

我应该如何使用正则表达式来查找和替换网址？我试图在这里弄明白：https://regex101.com/r/kF3gI1/1

Answer 1

所以我继续并假设所有条目都将使用双引号来包含图像URL。这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" />

需要用模板代码替换img src

1 个答案: