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我有一个 IDN 网址。如果我将它按原样放入 XMLReader 中,我会收到两个警告

Warning: XMLReader::open(http://www.джумла-тест.рф/master/article-category-blog?format=feed&type=rss): failed to open stream: HTTP request failed! in C:\Program Files (x86)\EasyPHP-12.1\www\joomla-cms2\libraries\joomla\feed\factory.php on line 44

Warning: XMLReader::open(): Unable to open source data in C:\Program Files (x86)\EasyPHP-12.1\www\joomla-cms2\libraries\joomla\feed\factory.php on line 44
**Error. Feed could not be retrieved.**

如果我 urlencode 我得到一个

Warning: XMLReader::open(): Unable to open source data in C:\Program Files (x86)\EasyPHP-12.1\www\joomla-cms2\libraries\joomla\feed\factory.php on line 44
**Error. Feed could not be retrieved.**

如果我将其更改为 punycode 它可以工作,但这是正确的方法吗?

更新:我想补充一点,我认为问题可能更早,因为查询没有返回正确的 idn url。但我仍然想知道 php 中的 xmlreader 是否可以处理 idn url,假设我可以正确地从数据库中获取它们。

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After some discussion with other developers, it seems like the answer is "Yes you need to send the request as punycode."

于 2013-05-26T00:18:00.833 回答