SCRAPY模拟登陆链接重定向问题

Question

scrapy模拟登陆不能抓取到数据
模拟登陆没有跳转到抓取页面
真实去登陆账号不能再次的登陆了
不知道问题出在哪里求大神解答

 # -*- coding: utf-8 -*-
import scrapy
from scrapy.http import Request, FormRequest
from scrapy.selector import Selector
from scrapy.conf import settings

class faxinSpider(scrapy.Spider):

    name = "faxin"

    allowed_domains = ["www.faxin.cn"]
    start_urls = ["http://www.faxin.cn/keyword/index.aspx"]

    headers = settings.get('HEADERS')

    def start_requests(self):
        return [Request(
                    url = "http://www.faxin.cn/login.aspx", 
                    meta = {'cookiejar' : 1},
                    headers = self.headers, 
                    callback=self.post_login
                      )]

    def post_login(self, response):
        viewstate = Selector(response).xpath('//input[@name="__VIEWSTATE"]/@value').extract()[0]
        self.logger.info("--viewstate--:%s", viewstate)
        return [FormRequest.from_response(
                    response, 
                    meta = {'cookiejar' : response.meta['cookiejar']},
                    headers = self.headers,
                    formdata = {
                        '__VIEWSTATE' : viewstate,
                        'WebUCHead_Special1$hiddIsLogin' : '0',
                        'user_name' : '****',
                        'user_password' : '****'
                            },
                    callback = self.pares_page,
                        )]

    def after_login(self, response):
        for url in self.start_urls :
            yield Request(url, 
                           headers = self.headers, 
                           meta = {'cookiejar' : response.meta['cookiejar']}, 
                           callback = self.pares_page)

    def pares_page(self, response):
        url_ = response.url
        self.logger.info("---url---: %s", url_)
#         a = Selector(response).xpath("//a[@class='login']/@title")
#         self.logger.info("---username---: %s", a)
        a_ = Selector(response).xpath("//div[@class='t_box']/ul[@class='clearfix']/li")
        for a in a_:
            print a.xpath("a/@href").extract()[0]

Answer 1

https://segmentfault.com/q/1010000002431559