【发布时间】:2020-09-22 15:08:03
【问题描述】:
我正在尝试创建一个脚本来提交表单并将结果返回给我。我可以从 URL 中提取表单信息,但无法更新表单的字段或获得响应。
我目前有:
import requests
from bs4 import BeautifulSoup as bs
url = 'https://dos.elections.myflorida.com/campaign-finance/contributions/'
response = requests.get(url)
soup = bs(response.text)
form_info = soup.find_all('action')
print(form_info[0]['action'])
哪个有效并返回:
'/cgi-bin/contrib.exe'
这个表单应该可以默认提交,所以我再试试:
session = requests.Session()
BASE_URL = 'https://dos.elections.myflorida.com'
headers = {'User-Agent': "Mozilla/5.0" , 'referer' :'{}/campaign-finance/contributions/'.format(BASE_URL)}
data = {'Submit' : 'Submit'}
res = session.post( '{}/cgi-bin/contrib.exe'.format(BASE_URL), data = data, headers = headers )
我收到 502 响应。由于this post.
https://dos.elections.myflorida.com/campaign-finance/contributions/
结果将我重定向到:
https://dos.elections.myflorida.com/cgi-bin/contrib.exe
SIM 的解决方案奏效了,谢谢!!
【问题讨论】:
标签: python html web-scraping python-requests