PHP 网址验证错误:http://https://example.com(又名 https://https://example.com) [英] PHP Url Validation Error: http://https://example.com (aka https://https://example.com)

问题描述

我有这个 url 正则表达式模式:

I had this url regex pattern in place:

$pattern = "@\b(https?://[^\s()<>\[\]\{\}]{1,".$max_length_allowed_for_each_url."}(?:\([\w\d]+\)|([^[:punct:]\s]|/)))@";

它似乎可以很好地验证我扔给它的任何 URL，直到我意识到 https://http://google.com(显然即使是 stackoverflow 也在考虑一个有效的 URL(它使该 URL 可点击，而不是我，虽然它确实删除了一个冒号)所以也许我运气不好?)一个有效的 URL，当它肯定不是时.

It seemed to work pretty well at validating any URL I threw at it, until I realized that https://http://google.com (apparently even stackoverflow is considering that a valid URL (it made that URL clickable, not me, although it did remove one of the colons) so perhaps I am out of luck?) was a valid URL, when it certainly is not.

我做了一些研究...发现我应该使用 filter_var 而不是正则表达式来验证 PHP URL 无论如何......并且很失望地意识到它也容易受到同样的验证问题的影响.

I did a little research... and learnt that I should be using filter_var instead of a regex for PHP URL validation anyways... and was disappointed to realize that it too is susceptible to this very same validation problem.

我可以轻松地征服它:

str_replace(array("https://http://","http://https://"), array("http://","https://"), $url);

但是……这似乎大错特错.

But... that just seems so wrong.

推荐答案

嗯，这是一个有效的 URI.从技术上讲.如果您不相信我，请查看 RFC 以获取 URI.

Well, it is a valid URI. Technically. Look at the RFC for URIs if you don't believe me.

• URI 的路径组件可以包含 //.
• http 是一个有效的主机名.
• 即使 : 存在(指定为 *digit，而不是 1*digit)，也允许缺少端口.(这就是 Stack Overflow 删除冒号的原因——它认为您使用的是默认端口，所以它从 URI 中删除了它.)

• The path component of a URI can contain //.
• http is a valid host name.
• The port is allowed to be missing even if the : is present (it's specified as *digit, not 1*digit). (This is why Stack Overflow removed the colon -- it thought you were using the default port, so it removed it from the URI.)

我建议为此写一个特例.在单独的步骤中，检查 URI 是否以 https?://https?:// 开头，然后修复它.

I suggest writing a special case for this. In a separate step, check to see if the URI starts with https?://https?://, and fix it.

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