Can't scrape an image url from Zara - web-scraping

Question

I am trying to scrape an image url from Zara, but the only think I get back is the url of the transparent background.
This is the link Im trying to scrape: https://static.zara.net/photos///2022/V/0/1/p/9598/176/406/2/w/850/9598176406_1_1_1.jpg?ts=1640187784252
This is the link I keep getting:
https://static.zara.net/stdstatic/1.249.0-b.13/images/transparent-background.png'
Any ideas? This is my code. Thank you in advance!!
*Note: I used extract() in the image, not extract_first(), to see if there were several links, but they are all the same.
import scrapy
from scrapy.linkextractors import LinkExtractor
from Zara.items import Producto
class ZaraSpider(scrapy.Spider):
name = 'zara'
allowed_domains = ['zara.com']
start_urls = [
'https://www.zara.com/es/es/jersey-punto-cuello-subido-p09598176.html'
]
def parse(self, response):
producto = Producto()
# Extraemos los enlaces
links = LinkExtractor(
allow_domains=['zara.com'],
restrict_xpaths=["//a"],
allow="/es/es/"
).extract_links(response)
outlinks = [] # Lista con todos los enlaces
for link in links:
url = link.url
outlinks.append(url) # Añadimos el enlace en la lista
yield scrapy.Request(url, callback=self.parse) # Generamos la petición
product = response.xpath('//meta[#content="product"]').extract()
if product:
# Extraemos la url, el nombre del producto, la descripcion y su precio
producto['url'] = response.request.url
producto['nombre'] = response.xpath('//h1[#class="product-detail-info__name"]/text()').extract_first()
producto['precio'] = response.xpath('//span[#class="price__amount-current"]/text()').extract_first()
producto['descripcion'] = response.xpath('//div[#class="expandable-text__inner-content"]//text()').extract_first()
producto['imagen'] = response.xpath('//img[#class="media-image__image media__wrapper--media"]/#src').extract()
#producto['links'] = outlinks
yield producto'''

Answer 1

So the problem that it's generated with javascript. Try to request a webpage with scrapy shell and view the response, then you'll see that you can find to requested image url in another way.
import scrapy
from scrapy.linkextractors import LinkExtractor
# from Zara.items import Producto
class Producto(scrapy.Item):
url = scrapy.Field()
nombre = scrapy.Field()
precio = scrapy.Field()
descripcion = scrapy.Field()
imagen = scrapy.Field()
links = scrapy.Field()
class ZaraSpider(scrapy.Spider):
name = 'zara'
allowed_domains = ['zara.com']
start_urls = [
'https://www.zara.com/es/es/jersey-punto-cuello-subido-p09598176.html'
]
def parse(self, response):
producto = Producto()
# Extraemos los enlaces
links = LinkExtractor(
allow_domains=['zara.com'],
restrict_xpaths=["//a"],
allow="/es/es/"
).extract_links(response)
outlinks = [] # Lista con todos los enlaces
for link in links:
url = link.url
outlinks.append(url) # Añadimos el enlace en la lista
yield scrapy.Request(url, callback=self.parse) # Generamos la petición
product = response.xpath('//meta[#content="product"]').get()
if product:
# Extraemos la url, el nombre del producto, la descripcion y su precio
producto['url'] = response.request.url
producto['nombre'] = response.xpath('//h1[#class="product-detail-info__name"]/text()').get()
producto['precio'] = response.xpath('//span[#class="price__amount-current"]/text()').get()
producto['descripcion'] = response.xpath('//div[#class="expandable-text__inner-content"]//text()').get()
producto['imagen'] = response.xpath('//meta[#property="og:image"]/#content').get()
#producto['links'] = outlinks
yield producto
BTW check out CrawlSpider.