Python查询阿里巴巴关键字排名的方法

时间:2022-09-18 20:18:28

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

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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

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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。