如何从 JavaScript 呈现的响应页面下载最高分辨率的图像？

Question

Suppose this is the website page: " https://www.dior.com/en_us/products/couture-943C105A4655_C679-technical-fabric-cargo-pants-covered-in-tulle ", from which I want to download all the images of the product showcased (4 images in this case).

I am using Selenium and extracting image links. The problem is if I click the images they are even 2000x3000 pixels big, but I am only able to get 480 around pixels resolution images of them. Where are these images stored? How do I extract them? ( basically I want to download the maximum possible size of those images )

Answer 1

Withing the source code of the page you provided, there is json data that provides the links and content for the page. Once the data is stripped from the script in the source code, it is easy to retrieve the high resolution links and download the image. If you have not already, pip install requests and pip install bs4 .

import requests, re, json
from bs4 import BeautifulSoup

url = 'https://www.dior.com/en_us/products/couture-943C105A4655_C679-technical-fabric-cargo-pants-covered-in-tulle'

r = requests.get(url)
soup = BeautifulSoup(r.text, 'lxml')
script = [script.text for script in soup.find_all('script') if 'window.initialState' in script.text][0]
json_data_s = re.search(r'{.+}', script).group(0)
json_data = json.loads(json_data_s)
for holder in json_data['CONTENT']['cmsContent']['elements']:
    if holder.get('type') == 'PRODUCTMEDIAS':
        for image in holder['items']:
            name = image['galleryImages']['imageZoom']['viewCode']
            img_src = image['galleryImages']['imageZoom']['uri']
            image_page = requests.get(img_src)
            with open(name + '.jpg', 'wb') as img:
                img.write(image_page.content)

*The images you were downloading before were the thumbnail photos.