我可以使用从 here 中找到的以下 python 代码登录 amazon.com:
import mechanize
br = mechanize.Browser()
br.set_handle_robots(False)
br.addheaders = [("User-agent", "Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.9.2.13) Gecko/20101206 Ubuntu/10.10 (maverick) Firefox/3.6.13")]
sign_in = br.open('https://www.amazon.com/gp/sign-in.html')
br.select_form(name="sign-in")
br["email"] = 'test@test.com'
br["password"] = 'test4test'
logged_in = br.submit()
orders_html = br.open("https://www.amazon.com/gp/css/history/orders/view.html?orderFilter=year-%s&startAtIndex=1000" % 2013)
但以下两段使用请求模块和 urllib2 不起作用。
import requests
import sys
username = "test@test.com"
password = "test4test"
login_data = ({
'email' : fb_username,
'password' : fb_password,
'flex_password': 'true'})
url = 'https://www.amazon.com/gp/sign-in.html'
agent ={'User-agent', 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_7_4) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.57 Safari/537.1'}
session = requests.session(config={'verbose': sys.stderr}, headers = agent)
r = session.get('http://www.amazon.com')
r1 = session.post(url, data=login_data, cookies=r.cookies)
r2 = session.post("https://www.amazon.com/gp/css/history/orders/view.html?orderFilter=year-2013&startAtIndex=1000", cookies = r1.cookies)
#
import urllib2
import urllib
import cookielib
amazon_username = "test@test.com"
amazon_password = "test4test"
url = 'https://www.amazon.com/gp/sign-in.html'
cookie = cookielib.CookieJar()
login_data = urllib.urlencode({'email' : amazon_username, 'password' : amazon_password,})
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cookie))
opener.addheaders = [('User-agent', 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_7_4) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.57 Safari/537.1')]
opener.open('www.amazon.com')
response = opener.open(url, login_data)
response = opener.open("https://www.amazon.com/gp/css/history/orders/view.html?orderFilter=year-%s&startAtIndex=1000" % 2013, login_data)
我在发布亚马逊登录表单时做错了什么?这是我第一次发布表格。感谢您的帮助。
我更喜欢使用 urllib2 或请求,因为我的所有其他代码都使用这两个模块。
此外,任何人都可以评论一下 mechanize、requests 和 urllib2 之间的速度性能,以及 mechanize 相对于其他两者的其他优势吗?
~~~~~~~~~~~新~~~~~~~~~~~~ 按照 C.C. 的指示,我现在可以使用 urllib2 登录。但是当我尝试对请求做同样的事情时,它仍然不起作用。谁能给我一个线索?
import requests
import sys
fb_username = "test@test.com"
fb_password = "xxxx"
login_data = ({
'email' : fb_username,
'password' : fb_password,
'action': 'sign-in'})
url = 'https://www.amazon.com/gp/sign-in.html'
agent ={'User-agent', 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_7_4) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.57 Safari/537.1'}
session = requests.session(config={'verbose': sys.stderr}, headers = agent)
r = session.get(url)
r1 = session.post('https://www.amazon.com/gp/flex/sign-in/select.html', data=login_data, cookies=r.cookies)
b = r1.text
最佳答案
关于您的 urllib2
方法,您遗漏了两件事。
首先,如果您查看 sign-in.html
的源代码,它表明
<form name="sign-in" id="sign-in" action="/gp/flex/sign-in/select.html" method="POST">
意味着表单应该提交给 select.html
。
其次,除了邮箱和密码,您还需要选择您是否是现有用户:
<input id="newCust" type="radio" name="action" value="new-user"...>
...
<input id="returningCust" type="radio" name="action" value="sign-in"...>
它应该看起来像这样:
import cookielib
import urllib
import urllib2
amazon_username = ...
amazon_password = ...
login_data = urllib.urlencode({'action': 'sign-in',
'email': amazon_username,
'password': amazon_password,
})
cookie = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cookie))
opener.addheaders = [('User-agent', ...)]
response = opener.open('https://www.amazon.com/gp/sign-in.html')
print(response.getcode())
response = opener.open('https://www.amazon.com/gp/flex/sign-in/select.html', login_data)
print(response.getcode())
response = opener.open("https://www.amazon.com/") # it should show that you are logged in
print(response.getcode())
关于python - 为什么我可以使用 python mechanize 登录亚马逊网站,但不能使用 requests 或 urllib2,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/18265376/