如何安全地从URL获取文件扩展名? [英] How to safely get the file extension from a URL?

问题描述

考虑以下网址

http://m3u.com/tunein.m3u
http://asxsomeurl.com/listen.asx:8024
http://www.plssomeotherurl.com/station.pls?id=111
http://22.198.133.16:8024

确定文件扩展名(.m3u/.asx/.pls)的正确方法是什么?显然最后一个没有文件扩展名.

Whats the proper way to determine the file extensions (.m3u/.asx/.pls)? Obviously the last one doesn't have a file extension.

我忘了提到m3u/asx/pls是音频流的播放列表(文本文件)，必须以不同的方式进行解析.目标确定扩展名，然后将URL发送到适当的解析功能.例如.

I forgot to mention that m3u/asx/pls are playlists (textfiles) for audio streams and must be parsed differently. The goal determine the extension and then send the url to the proper parsing-function. E.g.

url = argv[1]
ext = GetExtension(url)
if ext == "pls":
  realurl = ParsePLS(url)
elif ext == "asx":
  realurl = ParseASX(url)
(etc.)
else:
  realurl = url
Play(realurl)

GetExtension()应该返回文件扩展名(如果有)，最好不要连接到URL.

GetExtension() should return the file extension (if any), preferrably without connecting to the URL.

推荐答案

real 正确的方法是根本不使用文件扩展名.对相关网址执行GET(或HEAD)请求，然后使用返回的"Content-type" HTTP标头获取内容类型.文件扩展名不可靠.

The real proper way is to not use file extensions at all. Do a GET (or HEAD) request to the URL in question, and use the returned "Content-type" HTTP header to get the content type. File extensions are unreliable.

请参见 MIME类型(IANA媒体类型)有关更多信息和有用的MIME类型的列表.

See MIME types (IANA media types) for more information and a list of useful MIME types.

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