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Елена Сергеева
Asked:2022-07-01 20:22:17 +0000 UTC2022-07-01 20:22:17 +0000 UTC

如何从url中删除部分参数

  • 772

我使用 urllib 获取参数：

from urllib import parse

for url in urls:
    all_instances = parse.urlparse(url).query
    dict_from_query = parse.parse_qs(query)

有一个不必要的参数列表：

qw = ["perpage", "h"]

以及链接列表：

urls = ["google.com/news?page=1&perpage=20", "google.com/news?h=34748fdg&page=20", "google.com/news?h=jshjdfh5&perpage=20"]

如何从链接中删除不必要的参数以得到以下列表：

urls = ["google.com/news?page=1", "google.com/news?page=20", "google.com/news"]
python

2 个回答

  1. 因为问题是关于正则表达式的，所以这是我的版本，难看，但它似乎有效：

    from re import sub

qw = ["perpage", "h"]
urls = ["google.com/news?page=1&perpage=20", 
        "google.com/news?h=34748fdg&page=20", 
        "google.com/news?h=jshjdfh5&perpage=20"]

for i in range(len(urls)):
    urls[i] = sub(fr'&?(?:{"|".join(qw)})=.*?(?=&|$)&?', '', urls[i]).rstrip('?')

print(urls)  # ['google.com/news?page=1', 'google.com/news?page=20', 'google.com/news']
    • 1
  2. Best Answer

    在这种情况下，最好使用现成的解决方案。

    for i in range(len(urls)):
    url = parse.urlparse(urls[i])
    # получаем query в виде списка кортежей
    query = parse.parse_qsl(url.query)
    newQuery = dict()
    for k, v in query:
        # берем только нужные
        if k not in qw:
            newQuery[k] = v
    # строим новый урл
    # Результат urlparse не позволяет обновлять атрибуты,
    # но можно создать копию с измененными данными
    newUrl = url._replace(query=parse.urlencode(newQuery))
    urls[i] = parse.urlunparse(newUrl)

print(urls)  #  ['google.com/news?page=1', 'google.com/news?page=20', 'google.com/news']
    • 1

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