刮网页包含锚标记 using scrapy

Question

I am scraping manulife刮网页包含锚标记<a href = "#"> using scrapy

我想进入下一个页面，当我检查了“下一个”我得到：

<span class="pagerlink"> 
    <a href="#" id="next" title="Go to the next page">Next</a> 
</span> 

还有什么是正确的做法遵循？

# -*- coding: utf-8 -*- 
import scrapy 
import json 
from scrapy_splash import SplashRequest 

class Manulife(scrapy.Spider): 
name = 'manulife' 
#allowed_domains = ['https://manulife.taleo.net/careersection/external_global/jobsearch.ftl?lang=en'] 
start_urls = ['https://manulife.taleo.net/careersection/external_global/jobsearch.ftl?lang=en&location=1038'] 

def start_requests(self): 
    for url in self.start_urls: 
     yield SplashRequest(
     url, 
     self.parse, 
     args={'wait': 5}, 
     ) 




def parse(self, response): 
    #yield { 
    # 'demo' : response.css('div.absolute > span > a::text').extract() 
    #  } 
    urls = response.css('div.absolute > span > a::attr(href)').extract() 
    for url in urls: 
     url = "https://manulife.taleo.net" + url 
     yield SplashRequest(url = url, callback = self.parse_details, args={'wait': 5}) 
     #self.log("reaced22 : "+ url) 

    #hitting next button 
    #data = json.loads(response.text) 
    #self.log("reached 22 : "+ data) 
    #next_page_url = 

    if next_page_url: 
     next_page_url = response.urljoin(next_page_url) 
     yield SplashRequest(url = next_page_url, callback = self.parse, args={'wait': 5}) 


def parse_details(self,response): 
    yield { 
      'Job post' : response.css('div.contentlinepanel > span.titlepage::text').extract(), 
      'Location' : response.xpath("//span[@id = 'requisitionDescriptionInterface.ID1679.row1']/text()").extract(), 
      'Organization' : response.xpath("//span[@id = 'requisitionDescriptionInterface.ID1787.row1']/text()").extract(), 
      'Date posted' : response.xpath("//span[@id = 'requisitionDescriptionInterface.reqPostingDate.row1']/text()").extract(), 
      'Industry': response.xpath("//span[@id = 'requisitionDescriptionInterface.ID1951.row1']/text()").extract() 
      } 

正如您所看到的，代码包含SplashRequest，同时打到下一页的链接。

我是刮菜的新手，在某处我发现该网站可以返回响应为json也。我试过了，但它给我的错误，“没有JSON对象可以解码”

Answer 1

我认为使用CSS选择器".pagerlink a[title='Go to the next page']"这样可以工作。

但".pagerlink:last-child a"将是最好的方法。你只需要获得href属性