使用Python请求登录Instagram

Question

我试图在python中编写一个脚本,登录http://insta.friendorfollow.com/并获取没有关注的人员列表.我想使用"请求模块",到目前为止我做了很多尝试,没有运气.我的代码如下:

import requests, re

f = open('file', 'w')
r = requests.get('http://insta.friendorfollow.com/')
next_url = re.findall(ur'<a href=\"(.*)\" type=\"submit\"', r.content)

r = requests.get(next_url[0])

action = re.findall(ur'action=\"(.*)\"', r.content)
csrfmiddlewaretoken = re.findall(ur'name=\"csrfmiddlewaretoken\" value=\"(.*)\"', r.content)

print action
print csrfmiddlewaretoken

payload = {
'csrfmiddlewaretoken': csrfmiddlewaretoken[0],
'username': 'SOMEUSER',
'password':'SOMEPASS'
}

g = requests.post("https://instagram.com/"+action[0],
  data=payload, allow_redirects=True)


print >> f, g.text

有人可以告诉我我做错了什么吗？什么是正确的方式来做到这一点.一个脚本将非常感激.

Answer 1

一切都很好,

#!/usr/bin/env python

username = "username"
password = "password"

from selenium import webdriver
from selenium.webdriver.common.keys import Keys
import re

driver = webdriver.PhantomJS()
driver.get("http://insta.friendorfollow.com")
driver.find_elements_by_tag_name("a")[1].click()

print "Perimene file....".upper()
driver.find_element_by_name('username').send_keys(username)
driver.find_element_by_name('password').send_keys(password)
driver.find_element_by_class_name("button-green").click()

try:
  driver.find_elements_by_name("allow")[1].click()
except:
  pass

f = open(username+".txt", 'w')
malakes = re.findall(ur'data-id=\"([0-9]*)\"', driver.page_source)[::-1]
for malakas in malakes:
  print >> f, malakas

f.close()
driver.quit()