Image http://i.imgur.com/OigSBjF.png
import requests from bs4 import BeautifulSoup
r = requests.get("xxxxxxxxx")
soup = BeautifulSoup(r.content)
for link in links:
if "http" in link.get('src'):
print link.get('src')
I get the printed URL but don't know how to work with it.
You need to download and write to disk:
import requests
from os.path import basename
r = requests.get("xxx")
soup = BeautifulSoup(r.content)
for link in links:
if "http" in link.get('src'):
lnk = link.get('src')
with open(basename(lnk), "wb") as f:
f.write(requests.get(lnk).content)
You can also use a select to filter your tags to only get the ones with http links:
for link in soup.select("img[src^=http]"):
lnk = link["src"]
with open(basename(lnk)," wb") as f:
f.write(requests.get(lnk).content)
While the other answers are perfectly correct.
I found it really slow to download and don't know the progress with really high resolution images.
So, made this one.
from bs4 import BeautifulSoup
import requests
import subprocess
url = "https://example.site/page/with/images"
html = requests.get(url).text # get the html
soup = BeautifulSoup(html, "lxml") # give the html to soup
# get all the anchor links with the custom class
# the element or the class name will change based on your case
imgs = soup.findAll("a", {"class": "envira-gallery-link"})
for img in imgs:
imgUrl = img['href'] # get the href from the tag
cmd = [ 'wget', imgUrl ] # just download it using wget.
subprocess.Popen(cmd) # run the command to download
# if you don't want to run it parallel;
# and wait for each image to download just add communicate
subprocess.Popen(cmd).communicate()
Warning: It won't work on win/mac as it uses wget.
Bonus: You can see the progress of each image if you are not using communicate.
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