用于从Content-Disposition标头中提取文件名的javascript正则表达式

Question

Content-disposition标头包含可以轻松提取的文件名,但有时它包含双引号,有时没有引号,也可能还有其他一些变体.有人可以编写一个适用于所有情况的正则表达式.

Content-Disposition: attachment; filename=content.txt

以下是一些可能的目标字符串:

attachment; filename=content.txt
attachment; filename*=UTF-8''filename.txt
attachment; filename="EURO rates"; filename*=utf-8''%e2%82%ac%20rates
attachment; filename="omáèka.jpg"
and some other combinations might also be there

Answer 1

你可以尝试这种精神:

filename[^;=\n]*=((['"]).*?\2|[^;\n]*)

filename      # match filename, followed by
[^;=\n]*      # anything but a ;, a = or a newline
=
(             # first capturing group
    (['"])    # either single or double quote, put it in capturing group 2
    .*?       # anything up until the first...
    \2        # matching quote (single if we found single, double if we find double)
|             # OR
    [^;\n]*   # anything but a ; or a newline
)

您的文件名在第一个捕获组中:http://regex101.com/r/hJ7tS6

Answer 2

稍作修改以匹配我的用例（将所有引号和UTF标签剥离）

filename\*?=['"]?(?:UTF-\d['"]*)?([^;\r\n"']*)['"]?;?

https://regex101.com/r/UhCzyI/3

Answer 3

/filename[^;=\n]*=(?:(\\?['"])(.*?)\1|(?:[^\s]+'.*?')?([^;\n]*))/i

https://regex101.com/r/hJ7tS6/51

编辑:您也可以使用此解析器:https: //github.com/Rob--W/open-in-browser/blob/master/extension/content-disposition.js

Answer 4

filename[^;\\n]*=(UTF-\\d[\'"]*)?(([\'"]).*?[.]$\\2|[^;\\n]*)?\n

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我已经升级了Robin\xe2\x80\x99s 解决方案以执行另外两件事：

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1. 捕获文件名，即使它已转义双引号。\n

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2. 将 UTF-8\'\' 部分捕获为单独的组。\n

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这是一个 ECMAScript 解决方案。

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https://regex101.com/r/7Csdp4/3/

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