python - 无法在python中使用scrapy迭代数据

Question

我正在下面的网站上抓取女 Actor 姓名、排名和分数 http://www.timescelebex.com/top_actresses

我可以获得 1 条记录，但无法使用 for 迭代查找其余记录

我可以接收如下数据，但只有 1 条记录

{'分数':u'41.0'，'姓名':u'Deepika Padukone'，'排名':u'1'}

我想像上面那样提取所有记录

# -*- coding: utf-8 -*-
import scrapy


class ActressListSpider(scrapy.Spider):
    name = 'actress_list'
    allowed_domains = ['timescelebex.com/top_actresses']
    start_urls = ['http://timescelebex.com/top_actresses/']

    def parse(self, response):
        names=response.xpath('/html/body/div[2]/section/div/div/div[1]/table/tbody/tr/td/table/tbody/tr/td[1]/table/tbody')

        for name in names:
            actress = name.xpath('//*[@class="Droid Ctxt1"]/text()').extract_first()
            rank = name.xpath('//*[@class="Droid Stext"]/text()').extract_first()
            score = name.xpath('//*[@class="Stext2"]/text()').extract_first()

            yield{'Name':actress,'Rank':rank,'Score':score}

如果我使用 .extract 而不是 .extract_first() 我可以看到所有数据，但其输入如下

{'得分': , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ]>, '名称': , , , , , , , , , , , , , , , , , , ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,] >, ‘排名’: , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,]>}

Answer 1

您需要在 for 循环中使用相对 XPath:

def parse(self, response):
    names=response.xpath('//table[@class="ITable"]//table//tr[position() > 2]')

    for name in names:
        actress = name.xpath('.//*[@class="Droid Ctxt1"]/text()').extract_first()
        rank = name.xpath('.//*[@class="Droid Stext"]/text()').extract_first()
        score = name.xpath('.//*[@class="Stext2"]/text()').extract_first()

        yield{'Name':actress,'Rank':rank,'Score':score}