从URL的Python的XPath返回空列表

Question

我试图访问使用XPath以下URL的元素： http://www.booking.com/searchresults.html?dest_id=2400&dest_type=region&offset=288

我正在寻找的特定元素是div类“sr_item_link_to_villas”。我一直在使用以下XPath试图访问它（在这个例子中，我试图访问第二上市，但完整的脚本通过每个上市循环），但它返回一个空列表：

//*[@id="hotellist_inner"]/*[contains(@class,"sr_item")][2]//*[contains(@class,"sr_item_link_to_villas ")] 

的完整的代码是：

url='http://www.booking.com/searchresults.html?dest_id=2400&dest_type=region&offset=288' 
page = parse(url).getroot() 
pathstr='//*[@id="hotellist_inner"]/*[contains(@class,"sr_item")][2]//*[contains(@class,"sr_item_link_to_villas ")]' 
content=page.xpath(pathstr) 

Answer 1

以下代码可能会解决您的目的。您必须为获取数据添加标题值。

import urllib2 
    from lxml import etree 
    from lxml.html import tostring,fromstring 

    def get_HTML(url): 
     header={"User-Agent": "Mozilla/5.0 (Windows NT 6.1; WOW64; rv:41.0) Gecko/20100101 Firefox/41.0","Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8","Connection": "keep-alive"} 
     req=urllib2.Request(url,None,header) 
     return urllib2.urlopen(req).read() 

    url="http://www.booking.com/searchresults.html?dest_id=2400&dest_type=region&offset=288" 

    read = get_HTML(url) 
    tree = etree.HTML(read)  
    data = tree.xpath("//div[@class='sr_item_link_to_villas ']/a/text()"); 
    print data