如何将 UTF-8 中的网络抓取图像 link 编码为 ASCII 但仍然具有功能 link?

How to encode a webscraped image link in UTF-8 to ASCII but still have a functional link?

我正在尝试将 link 抓取到图像中,以便在我的 Kivy 应用程序中使用它。问题是图像地址中有波兰语符号(ę、ł、ó、ą),我收到此错误:

UnicodeEncodeError: 'ascii' codec can't encode characters in position 36-37: ordinal not in range(128)

完整的错误回溯:

Traceback (most recent call last):
  File "F:\Kivy\lib\site-packages\kivy\loader.py", line 342, in _load_urllib
    fd = opener.open(request)
  File "c:\users\user\appdata\local\programs\python\python36\lib\urllib\request.py", line 526, in open
    response = self._open(req, data)
  File "c:\users\user\appdata\local\programs\python\python36\lib\urllib\request.py", line 544, in _open
    '_open', req)
  File "c:\users\user\appdata\local\programs\python\python36\lib\urllib\request.py", line 504, in _call_chain
    result = func(*args)
  File "c:\users\user\appdata\local\programs\python\python36\lib\urllib\request.py", line 1361, in https_open
    context=self._context, check_hostname=self._check_hostname)
  File "c:\users\user\appdata\local\programs\python\python36\lib\urllib\request.py", line 1318, in do_open
    encode_chunked=req.has_header('Transfer-encoding'))
  File "c:\users\user\appdata\local\programs\python\python36\lib\http\client.py", line 1239, in request
    self._send_request(method, url, body, headers, encode_chunked)
  File "c:\users\user\appdata\local\programs\python\python36\lib\http\client.py", line 1250, in _send_request
    self.putrequest(method, url, **skips)
  File "c:\users\user\appdata\local\programs\python\python36\lib\http\client.py", line 1117, in putrequest
    self._output(request.encode('ascii'))
UnicodeEncodeError: 'ascii' codec can't encode character '\u0142' in position 36: ordinal not in range(128)
[INFO   ] [GL          ] NPOT texture support is available
[INFO   ] [WindowSDL   ] exiting mainloop and closing.
[INFO   ] [Base        ] Leaving application in progress...

Process finished with exit code 0

这里有一个例子,你可以明白我的意思。在正常加载图片时,没有错误,第二个输出 UnicodeEncodeError 并显示黑色。

from kivy.app import App
from kivy.lang import Builder

build_structure = """
Screen:
    BoxLayout:
        AsyncImage:
            # This doesnt load because it's in UTF-8 and outputs the error above 
            # but it doesn't break the app.

            source: app.link_to_image_bad
        AsyncImage:
            # This one does load
            source: app.link_to_image_good
"""


class ImageApp(App):
    # This link has Polish signs in it so it will give the UnicodeEncodeError
    link_to_image_bad = "https://nowa.1lo.gorzow.pl/wp-content/uploads/2020/11/Szkoła-do-hymnu.png"

    link_to_image_good = "https://nowa.1lo.gorzow.pl/wp-content/uploads/2020/11/Olimpiada-statystyczna.png"

    def build(self):
        return Builder.load_string(build_structure)


if __name__ == '__main__':
    ImageApp().run()

以上代码的输出:

有没有办法避免这个错误并且仍然可以正常工作link?

URL 应该已经是 ASCII 兼容的。互联网(又名 HTTP)上的流量是这样工作的:只有 ASCII URLS(有额外的限制)。浏览器现在倾向于取消转义 URL。 [我们在 URL 中部分看到的 %20 和其他 %xx 字符。注意:现在我们有 UTF-8 编码,并且最重要的是 URL 转义。所以,你应该记住你有两层编码。

你应该逃脱 URL,见 URL quoting。我会使用 quote()unquote()。在评论中,我们看到了一个quote_plus(),但那个也改变了space,有时间用,但它会改变原始数据的含义。

编辑:

好的,我有问题。 kivy 如何处理 URLS 似乎有些奇怪。 quote() 仅用于路径部分,不适用于 URL.

的第一部分

作为 hack(如果你有一个特定的端口,它就不起作用:它会在端口前面引用 :):

url = 'https://nowa.1lo.gorzow.pl/wp-content/uploads/2020/11/Szkoła-do-hymnu.png'
url_split = url.split('//')
'//'.join([url_split[0], urllib.parse.quote(url_split[1]))

所以你得到了想要的:'https://nowa.1lo.gorzow.pl/wp-content/uploads/2020/11/Szko%C5%82a-do-hymnu.png' 浏览器使用。

您可能希望将它包含在您自己的函数中(并且可能检查是否有端口号,以将其排除在引号之外)。

等等,也许有人有 Kivy 的真正解决方案。我从不使用完全限定路径(协议和域也是如此),所以对我来说基本 quote() 就足够了。