How to get full image url with python

Question

I have some problem with parsing links for images (images of house) on this site (https://kvartiry-bolgarii.ru/trekhkomnatnaya-kvartira-v-blagoustroennom-i-spokoynom-kurortnom-poselke-o26252)

How can i get full link? How can I get data from src (in all images) and combine it into full link with site domain?

Im try it but cant get full link because dont kbow how to take link in src

import requests
from bs4 import BeautifulSoup

rs = requests.get('https://kvartiry-bolgarii.ru/neveroyatnaya-kvartira-s-vidom-na-more-tip-pentkhaus-o26253')
root = BeautifulSoup(rs.content, 'html.parser')

urls = root.select('#slider > li > img[src]')
print(urls)
# [<img alt="" src="/photos/5e2c79b4-7da2-478e-a783-ad8f010d0b15.jpg"/>, , <img alt="" src="/photos/90f58624-1f32-46a2-afc9-ad8f010e2703.jpg"/>]

Do you want the image URLs shown in slider? Can you use beautifulsoup? — Andrej Kesely
– Andrej Kesely, Commented Aug 30, 2021 at 17:50
Let me know if you are looking for solution using Selenium ? — cruisepandey
– cruisepandey, Commented Aug 30, 2021 at 18:23

Tim Roberts · Accepted Answer · 2021-08-30 18:14:18Z

1

I don't understand how you could get that far and not know how to get the src attribute:

import requests
from bs4 import BeautifulSoup

base = 'https://kvartiry-bolgarii.ru/neveroyatnaya-kvartira-s-vidom-na-more-tip-pentkhaus-o26253'

rs = requests.get(base)
root = BeautifulSoup(rs.content, 'html.parser')

urls = root.select('#slider > li > img[src]')
for url in urls:
    print( base+url['src'] )

answered Aug 30, 2021 at 18:14

Tim Roberts

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