python请求url路径拼接parse用法

# encoding: utf-8
"""
@time: 2021/9/1 13:48
"""
from urllib.parse import urlparse, urlunparse, parse_qs, urlencode


def replace_field(url, name, value):
    parse = urlparse(url)
    query = parse.query  # 对象的.query 属性，是一个字符串，也就是网址中，问号后面的内容
    query_pair = parse_qs(query)
    # parse_qs与urlencode也是一对相反函数。其中前者把 .query输出的字符串转成字典，
    # 而后者把字段转成.query形式的字符串
    query_pair[name] = value
    new_query = urlencode(query_pair, doseq=True)
    new_parse = parse._replace(query=new_query)
    next_page = urlunparse(new_parse)
    return next_page


if __name__ == '__main__':
    url_list = [
        'https://xxx.com/articlelist?category=technology',
        'https://xxx.com/articlelist?category=technology&after=',
        'https://xxx.com/articlelist?category=technology&after=asdrtJKSAZFD',
        'https://xxx.com/articlelist?category=technology&after=asdrtJKSAZFD&other=abc'
    ]

    for url in url_list:
        next_page = replace_field(url, 'after', '0000000')
        print(next_page)
    # https://xxx.com/articlelist?category=technology&after=0000000
    # https://xxx.com/articlelist?category=technology&after=0000000
    # https://xxx.com/articlelist?category=technology&after=0000000
    # https://xxx.com/articlelist?category=technology&after=0000000&other=abc