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。