使用Xpath提取錨定標記的href給定文本

Question

我需要從給出錨定標記的特定文本的href屬性中提取url。

from scrapy.spider import Spider 
from scrapy.selector import Selector 
from nba.items import NBAItem 

class ESPNSpider(Spider): 
    name = "ESPN" 
    allowed_domains = ["espn.com"] 
    start_urls = ["http://espn.go.com/nba/teams"] 

def parse(self, response): 
    sel = Selector(response) 
    sites = sel.xpath('//*[@id="content"]/div[3]/div[1]') 
    items = [] 
    for site in sites: 
     item = NBAItem() 
     item['team_name'] = site.xpath('//a[@class="bi"]/text()').extract() 
     item['team_link'] = site.xpath('//a[@class="bi"]/@href').extract() 
     item['team_stats_link'] = site.xpath('//a[text()='Stats']/@href').extract() 
     items.append(item) 
    return items 

這是我遇到的麻煩線路：

item['team_stats_link'] = site.xpath('//a[text()='Stats']/@href').extract() 

我也嘗試：

item['team_stats_link'] = site.xpath('//a[contains(text(), 'Stats')]/@href).extract() 

相關網站：http://espn.go.com/nba/teams

Answer 1

你裏面的XPath循環應該以.//開頭，換句話說，您需要使其相關給site。

我還經歷li標籤ul與medium-logos類中，而不是與content ID的div內第三div裏面搜索第一個div：

class ESPNSpider(Spider): 
    name = "ESPN" 
    allowed_domains = ["espn.com"] 
    start_urls = ["http://espn.go.com/nba/teams"] 

    def parse(self, response): 
     sel = Selector(response) 
     sites = sel.xpath('//ul[@class="medium-logos"]//li') 
     for site in sites: 
      item = NBAItem() 
      item['team_name'] = site.xpath('.//a[@class="bi"]/text()').extract()[0] 
      item['team_link'] = site.xpath('.//a[@class="bi"]/@href').extract()[0] 
      item['team_stats_link'] = site.xpath(".//a[text()='Stats']/@href").extract()[0] 
      yield item 

它產生：

{'team_link': u'http://espn.go.com/nba/team/_/name/bos/boston-celtics', 'team_name': u'Boston Celtics', 'team_stats_link': u'/nba/teams/stats?team=bos'} 
{'team_link': u'http://espn.go.com/nba/team/_/name/bkn/brooklyn-nets', 'team_name': u'Brooklyn Nets', 'team_stats_link': u'/nba/teams/stats?team=bkn'} 
...