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。