使用 Python 保存访问 URL 时发送到浏览器的文件

Question

当我在浏览器中访问下面的 URL 时，它会自动下载一个 CSV 文件。由于内容每天更新，所以我想写一个Python命令每次获取最新的文件。

我试过 wget、requests 和 urllib.request - 都没有成功。

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

Answer 1

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)

内容为bytes对象，需要以二进制方式打开文件

使用 Python 保存访问 URL 时发送到浏览器的文件

Using Python to save a file that is sent to browser when visiting URL

csv

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python-3.x