我怎样才能缩短 python 中的 URL 3

Question

import http.client
import urllib.parse

def unshorten_url(url):
    parsed = urllib.parse.urlparse(url)
    h = http.client.HTTPConnection(parsed.netloc)
    resource = parsed.path
    if parsed.query != "":
        resource += "?" + parsed.query
    h.request('HEAD', resource )
    response = h.getresponse()
    if response.status/100 == 3 and response.getheader('Location'):
        return unshorten_url(response.getheader('Location')) # changed to process chains of short urls
    else:
        return url

unshorten_url("http://data.europa.eu/esco/occupation/00030d09-2b3a-4efd-87cc-c4ea39d27c34")

输入将是： http://data.europa.eu/esco/occupation/00030d09-2b3a-4efd-87cc-c4ea39d27c34 #yes 同样返回。'

我需要 展开后输出 URL : https://ec.europa.eu/esco/portal/occupation?uri=http%3A%2F%2Fdata.europa.eu%2Fesco%2Foccupation%2F00030d09-2b3a-4efd-87cc-c4ea39d27c34&conceptLanguage=en&full=true#&uri=http://data.europa.eu/esco/occupation/00030d09-2b3a-4efd-87cc-c4ea39d27c34'

Answer 1

如您所见，我有两个 URL，一个是短 URL，这是我的输入，另一个是完整 URL，以实现所需的输出 URL 我从一组同类 URL 中识别出一个模式。然后我写了这段代码并实现了要求的输出。

my_url = "http://data.europa.eu/esco/occupation/00030d09-2b3a-4efd-87cc-c4ea39d27c34"
a="https://ec.europa.eu/esco/portal/occupationuri=http%3A%2F%2Fdata.europa.eu%2Fesco%2Foccupation%2F"
b = my_url.split("/")[-1]
URL = a+ b+ "&conceptLanguage=en&full=true#&uri=" + my_url

输出即；所需的完整 URL 是 URL。 URL = " https://ec.europa.eu/esco/portal/occupation?uri=http%3A%2F%2Fdata.europa.eu%2Fesco%2Foccupation%2F00030d09-2b3a-4efd-87cc-c4ea39d27c34&conceptLanguage=en&full=true#&uri=http://data.europa.eu/esco/occupation/00030d09-2b3a-4efd-87cc-c4ea39d27c34'"

我怎样才能缩短 python 中的 URL 3

How can I unshorten a URL in python 3

urllib

python-3.x

http.client