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VisualizaçõesHow to scrape JavaScript rendered data from a website using Scrapy?
Using Scrapy, I'm trying to scrape data of tag <script type="application/ld+json">....
import json
class TestSpider(scrapy.Spider):
name = 'content'
start_urls = ['https://www.maserati.com/us/en/models/ghibli']
def parse(self, response):
for content in response.xpath('(//script[@type="application/ld+json"])/text()'):
data = json.loads(content)
yield {
'name': data['name'],
}
next_page = response.css('li.next a::attr("href")').get()
if next_page is not None:
yield response.follow(next_page, self.parse)
However, I'm not getting the test1.jl file that I was expecting once after writing, scrapy runspider test_spider.py - O test1.jl in the terminal
I just want the name for a start to know how it works.
Image and website link for inspection are given below:
Image that shows the javascript tag and the name property inside that I want to yield
Image of my code and the code in the terminal
·
Juan Pablo Isaza
1 Respostas
Responde à pergunta0
You were so close...just missing getall()
import scrapy
import json
class TestSpider(scrapy.Spider):
name = 'content'
start_urls = ['https://www.maserati.com/us/en/models/ghibli']
def parse(self, response):
for content in response.xpath('(//script[@type="application/ld+json"])/text()').getall():
data = json.loads(content)
yield {
'name': data['name'],
}
next_page = response.css('li.next a::attr("href")').get()
if next_page is not None:
yield response.follow(next_page, self.parse)
(I don't see any "next" button though)
·
Juan Pablo Isaza
Relatório
Responde à pergunta