抓取包含锚标记的网页 <a href = "#>使用scrapy [英] scraping web page containing anchor tag <a href = "#"> using scrapy
问题描述
我正在抓取 manulife
我想转到下一页,当我检查下一页"时,我得到:
I want to go to the next page, when I inspect the "next" I get :
<span class="pagerlink">
<a href="#" id="next" title="Go to the next page">Next</a>
</span>
应该遵循的正确方法是什么?
What could be the right approach to follow?
# -*- 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()
}
如您所见,该代码包含点击下一页链接时的 SplashRequest.
As you can see, the code contains the SplashRequest while hitting the next page link.
我是抓取的新手,在某处我发现该网站也可以将响应作为 json 返回.我试过了,但它给我的错误是无法解码 json 对象"
I am novice in scraping, somewhere I found that website can return the response as json also. I tried it , but it is giving me the error that " No json object could be decoded"
推荐答案
我认为使用 css 选择器 ".pagerlink a[title='Go to the next page']"
像这样可以工作.
I think using css selector ".pagerlink a[title='Go to the next page']"
like this could work.
但是 ".pagerlink:last-child a"
将是最好的方法.你只需要获取 href 属性
But ".pagerlink:last-child a"
would be the best approach imo. You just have to get the href attribute
这篇关于抓取包含锚标记的网页 <a href = "#>使用scrapy的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!