主页 / 问题 / 1370751
Accepted
Елена Сергеева
Asked:2022-06-12 18:46:38 +0000 UTC2022-06-12 18:46:38 +0000 UTC

如何从url获取参数？

  • 772

有一个链接：

google.com/news?page=1&perpage=20

如何从此 url 获取参数列表：

['page=1', 'perpage=20']
python

2 个回答

  1. Best Answer

    我会告诉你如何使用评论中的建议urllib.parse

    from urllib import parse
url = "google.com/news?page=1&perpage=20"

all_instances = parse.urlparse(url)

    该变量all_instances将包含链接中存在的所有实体

    >>> print(all_instances)
ParseResult(scheme='', netloc='', path='google.com/news', params='', query='page=1&perpage=20', fragment='')

    现在您可以访问参数query并获取必要的数据

    query = all_instances.query

    并解析值

    dict_from_query = parse.parse_qs(query) # Здесь будет словарь 
list_from_query = parse.parse_qsl(query) # Здесь будет список

print(f"Словарь = {dict_from_query}, список = {list_from_query}" )
# out
# Словарь = {'page': ['1'], 'perpage': ['20']}, список = [('page', '1'), ('perpage', '20')]

    或者你指定的那种

    list_from_query = query.split("&")

print(list_from_query)

#out 
# ['page=1', 'perpage=20']
    • 2 
  2. from w3lib.url import url_query_parameter
parms = []
url = "google.com/news?page=1&perpage=20"
parms.append(url_query_parameter(url, 'page'))
parms.append(url_query_parameter(url, 'perpage'))
print(parms)
    • 1

相关问题