将 src 属性值从相对 URL 替换为绝对 URL
这是一个简单的 preg_replace() 调用：
$string = 'src="index.php';
$string = preg_replace("/(src=('|\")+[^(http:|https:)])/i", "src=\"http://example.com/", $string);
echo $string;
我期望结果是 src="http://example.com/index.php 但结果是 src="http://example.com/ndex.php.
我一定在这里错过了一些东西..
Here's a simple preg_replace() call:
$string = 'src="index.php';
$string = preg_replace("/(src=('|\")+[^(http:|https:)])/i", "src=\"http://example.com/", $string);
echo $string;
I expect the result to be src="http://example.com/index.php but it turns out to be src="http://example.com/ndex.php.
I must be missing something here..
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这真是一个混乱的正则表达式。您到底想实现什么目标？看起来如果 URL 不是以 http 或 https 开头，您想要添加域吗？如果是这样，那么你就有点偏离了：
应该更接近目标。
这个正则表达式有什么作用？它查找：
src='或"http:< /code> 或https:注意：
{?!...)被称为否定前瞻 是零宽度断言 的一个示例。这里意味着它不消耗任何输入。在这种情况下，它意味着“后面没有...”。它会查找：
src='或"(http:|https:) 的任何一个字符（这就是[^...]构造的含义）注意：
相当于：
表示任何不<的字符/em> 这些字符之一。
That's a really messed up regex. What are you trying to achieve exactly? It looks like if the URL doesn't start with http or https you want to add the domain? If so, you're quite a bit off:
should be a lot closer to the mark.
What does this regex do? It looks for:
src='or"http:orhttps:Note:
{?!...)is called a negative lookahead and is one example of a zero-width assertion. "Zero-width" here means that it doesn't consume any of the input. In this case it means "not followed by ...".What does your regex do? It looks for:
src='or"characters(http:|https:)(that's what the[^...]construct means)Note:
is equivalent to:
meaning any character that is not one of those characters.
构造
[^(http:|https:)]不正确。它匹配除(,h,t,p,:,|、s或)。The construct
[^(http:|https:)]is incorrect. It matches any character except(,h,t,p,:,|,s, or).试试这个：我自己测试过
try this: I tested it myself