Python如何查询阿里巴巴的关键字排名?

Python如何查询阿里巴巴的关键字排名?

本文实例讲述了Python查询阿里巴巴关键字排名的方法。分享给大家供大家参考。具体如下:

这里使用python库urllib及pyquery基本东西的应用,实现阿里巴巴关键词排名的查询,其中涉及到urllib代理的设置,pyquery对html文档的解析

1. urllib 基础模块的应用,通过该类获取到url中的html文档信息,内部可以重写代理的获取方法

class ProxyScrapy(object):
 def __init__(self):
  self.proxy_robot = ProxyRobot()
  self.current_proxy = None
  self.cookie = cookielib.CookieJar()
 def __builder_proxy_cookie_opener(self):  
  cookie_handler = urllib2.HTTPCookieProcessor(self.cookie)
  handlers = [cookie_handler]
  if PROXY_ENABLE:
   self.current_proxy = ip_port = self.proxy_robot.get_random_proxy()
   proxy_handler = urllib2.ProxyHandler({'http': ip_port[7:]})
   handlers.append(proxy_handler)
  opener = urllib2.build_opener(*handlers)
  urllib2.install_opener(opener)
  return opener
 def get_html_body(self,url):
  opener = self.__builder_proxy_cookie_opener()
  request=urllib2.Request(url)
  #request.add_header("Accept-Encoding", "gzip,deflate,sdch")
  #request.add_header("Accept", "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8")
  #request.add_header("Cache-Control", "no-cache")
  #request.add_header("Connection", "keep-alive")
  try:
   response = opener.open(request,timeout=2)
   http_code = response.getcode()
   if http_code == 200:
    if PROXY_ENABLE:
     self.proxy_robot.handle_success_proxy(self.current_proxy)
    html = response.read()
    return html
   else:
    if PROXY_ENABLE:
     self.proxy_robot.handle_double_proxy(self.current_proxy)
    return self.get_html_body(url)
  except Exception as inst:
   print inst,self.current_proxy
   self.proxy_robot.handle_double_proxy(self.current_proxy)
   return self.get_html_body(url)

2. 根据输入的公司名及关键词列表,返回每个关键词的排名

def search_keywords_rank(keyword_company_name, keywords):
 def get_context(url):
  start=clock()
  html=curl.get_html_body(url)
  finish=clock()
  print url,(finish-start)
  d = pq(html)
  items = d("#J-items-content .ls-item")
  items_c = len(items)
  print items_c
  if items_c < 38:
   return get_context(url)
  return items, items_c
 result = OrderedDict()
 for keyword in keywords:
  for page_index in range(1,9):
   u = url % (re.sub('\s+', '_', keyword.strip()), page_index)
   items, items_c = get_context(u)
   b = False
   for item_index in range(0, items_c):
    e=items.eq(item_index).find('.title a')
    p_title = e.text()
    p_url = e.attr('href')
    e=items.eq(item_index).find('.cright h3 .dot-product')
    company_name = e.text()
    company_url = e.attr('href')
    if keyword_company_name in company_url:
     total_index = (page_index-1)*38 +item_index+1+(0 if page_index==1 else 5)
     print 'page %s, index %s, total index %s' % (page_index, item_index+1, total_index)
     b = True
     if keyword not in result:
      result[keyword] = (p_title, p_url, page_index, item_index+1, total_index, u)
     break
   if b:
    break
 return result

希望本文所述对大家的Python程序设计有所帮助。