"https://www.tokopedia.com/sitemap/product/1.xml.gz" this is my url this url contains the number of product urls but it's zipped i don't know how to unzip the url and how to get the data from that, how to unzip it using scrapy or Beautiful soup some other scrapy libraries
Take a look at gzip
import requests
from io import BytesIO
import gzip
r = requests.get('https://www.tokopedia.com/sitemap/product/1.xml.gz',stream=True)
g=gzip.GzipFile(fileobj=BytesIO(r.content))
content=g.read()
print(content)
Output is too long to be pasted here. So giving output for g.read(1000)
Output:
b'<?xml version="1.0" encoding="UTF-8"?>\n\t<urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"\n\txmlns:xhtml="http://www.w3.org/1999/xhtml">\n\t <url>\n\t <loc>https://www.tokopedia.com/tokoshishaonline/shisha-medium</loc>\n <xhtml:link rel="alternate" media="only screen and (max-width: 640px)" href="https://m.tokopedia.com/tokoshishaonline/shisha-medium" />\n\t </url>\n\t <url>\n\t <loc>https://www.tokopedia.com/lighting/lampu-sorot-philips-hnf-207-flood-light-lampu-tembak-lampu-stadion</loc>\n <xhtml:link rel="alternate" media="only screen and (max-width: 640px)" href="https://m.tokopedia.com/lighting/lampu-sorot-philips-hnf-207-flood-light-lampu-tembak-lampu-stadion" />\n\t </url>\n\t <url>\n\t <loc>https://www.tokopedia.com/agromedia/pop-supernasa</loc>\n <xhtml:link rel="alternate" media="only screen and (max-width: 640px)" href="https://m.tokopedia.com/agromedia/pop-supernasa" />\n\t </url>\n\t <url>\n\t <loc>https://www.tokopedia.com/agromedia/aero-810</loc>\n <xhtml:l'