Python 请求转到链接和下载

Question

我想以自动化方式执行以下操作：

1. 转到此链接：https://www.cms.gov/Research-Statistics-Data-and-Systems/Statistics-Trends-and-Reports/MCRAdvPartDEnrolData/Monthly-Enrollment-by-Contract-Plan-State-County-DL.xml

2. 点击页面最底部的链接（以当前年份和月份结尾（即http://www.cms.gov/Research-Statistics-Data-and-Systems/Statistics-Trends-and-Reports/MCRAdvPartDEnrolData/Monthly-Enrollment-by-Contract-Plan-State-County-Items/Monthly-Enrollment-by-CPSC-2016-04.html）

3. 在下一页，从“下载”下的顶部链接下载 zip 文件： CPSC 每月注册 - 2016 年 4 月 [ZIP, 20MB]

到目前为止，我有以下获取当前年份和月份的信息，但我需要其他帮助...

from datetime import datetime
import calendar
Day = datetime.now().day
Month = datetime.now().month
Year = datetime.now().year
m=calendar.month_name[Month]

Answer 1

您需要一个 XML 解析器来从 XML 提要中提取链接，并需要一个 HTML 解析器来提取到 zip 文件的链接。为此，我们将分别使用lxml.etree 和lxml.html。工作实施：

from datetime import datetime
from urllib.request import urlretrieve
from urllib.parse import urljoin

import requests
from lxml import etree
from lxml import html


date_part = datetime.now().strftime("%Y-%m")
with requests.Session() as session:
    # get the XML feed and extract the link
    response = session.get("https://www.cms.gov/Research-Statistics-Data-and-Systems/Statistics-Trends-and-Reports/MCRAdvPartDEnrolData/Monthly-Enrollment-by-Contract-Plan-State-County-DL.xml")
    root = etree.fromstring(response.content)
    link = root.xpath("//item/link[contains(., '-%s.html')]/text()" % date_part)[0]

    # follow the link and extract the link to the zip file
    response = session.get(link)
    root = html.fromstring(response.content)
    zip_link = root.xpath("//a[@type='application/zip']/@href")[0]
    link = urljoin(link, zip_link)

    # download zip
    urlretrieve(link, filename="my.zip")