将文件名转换为file://URL [英] Convert a filename to a file:// URL

问题描述

在WeasyPrint的公共API中，我接受HTML输入的文件名(除其他类型外).可以与内置open()一起使用的任何文件名都可以使用，但是我需要将其转换为file://方案中的URL，然后再将其传递给urllib.urlopen().

In WeasyPrint’s public API I accept filenames (among other types) for the HTML inputs. Any filename that works with the built-in open() should work, but I need to convert it to an URL in the file:// scheme that will later be passed to urllib.urlopen().

(所有内容在内部都是URL形式.为了使用urlparse.urljoin()解析相对URL引用，我需要文档的基本URL".)

(Everything is in URL form internally. I need to have a "base URL" for documents in order to resolve relative URL references with urlparse.urljoin().)

urllib.pathname2url 是一个开始:

将路径名路径从路径的本地语法转换为URL路径组件中使用的形式. 这不会产生完整的URL.返回值已经使用quote()函数进行了引用.

Convert the pathname path from the local syntax for a path to the form used in the path component of a URL. This does not produce a complete URL. The return value will already be quoted using the quote() function.

重点是我的，但我确实需要完整的URL.到目前为止，这似乎可行:

The emphasis is mine, but I do need a complete URL. So far this seems to work:

def path2url(path):
    """Return file:// URL from a filename."""
    path = os.path.abspath(path)
    if isinstance(path, unicode):
        path = path.encode('utf8')
    return 'file:' + urlparse.pathname2url(path)

UTF-8似乎是 RFC 3987(IRI)推荐的.但是在这种情况下(URL最终将用于urllib)也许我应该使用 sys.getfilesystemencoding()?

UTF-8 seems to be recommended by RFC 3987 (IRI). But in this case (the URL is meant for urllib, eventually) maybe I should use sys.getfilesystemencoding()?

但是，根据文献，我不仅应该在file:之前，而且应该在file://之前... ...我不应该:在Windows上，nturl2path.pathname2url()的结果已经以三个斜杠开头.

However, based on the literature I should prepend not just file: but file:// ... except when I should not: On Windows the results from nturl2path.pathname2url() already start with three slashes.

所以问题是:有没有更好的方法来做到这一点并使它跨平台?

So the question is: is there a better way to do this and make it cross-platform?

推荐答案

出于完整性考虑，在Python 3.4+中，您应该执行以下操作:

For completeness, in Python 3.4+, you should do:

import pathlib

pathlib.Path(absolute_path_string).as_uri()

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