使用Python保存访问URL时发送到浏览器的文件

如何解决使用Python保存访问URL时发送到浏览器的文件

当我在浏览器中访问以下 URL 时，它会自动下载 CSV。内容每天更新，想写个python命令每次都获取最新的文件。

我尝试过 wget、requests 和 urllib.request - 都没有运气。

url = 'https://coronavirus.data.gov.uk/api/v1/data?filters=areaType=overview&structure=%7B%22areaType%22:%22areaType%22,%22areaName%22:%22areaName%22,%22areaCode%22:%22areaCode%22,%22date%22:%22date%22,%22newPeopleVaccinatedFirstDoseByPublishDate%22:%22newPeopleVaccinatedFirstDoseByPublishDate%22,%22cumPeopleVaccinatedFirstDoseByPublishDate%22:%22cumPeopleVaccinatedFirstDoseByPublishDate%22%7D&format=csv'

有人有什么想法吗？ TIA

解决方法

这对我来说很好用：

import requests

url = 'https://coronavirus.data.gov.uk/api/v1/data?filters=areaType=overview&structure=%7B%22areaType%22:%22areaType%22,%22areaName%22:%22areaName%22,%22areaCode%22:%22areaCode%22,%22date%22:%22date%22,%22newPeopleVaccinatedFirstDoseByPublishDate%22:%22newPeopleVaccinatedFirstDoseByPublishDate%22,%22cumPeopleVaccinatedFirstDoseByPublishDate%22:%22cumPeopleVaccinatedFirstDoseByPublishDate%22%7D&format=csv'

r = requests.get(url)

with open("uk_data.csv","wb") as f:
    f.write(r.content)

内容是一个bytes对象，所以需要以二进制方式打开文件。