How do I make it so that each image I garnered from web scraping is then stored to a folder? I use Google Colab currently since I am just practicing stuff. I want to store them in my Google Drive folder.
This is my code for web scraping:
import requests
from bs4 import BeautifulSoup
def getdata(url):
r = requests.get(url)
return r.text
htmldata = getdata('https://www.yahoo.com/')
soup = BeautifulSoup(htmldata, 'html.parser')
imgdata = []
for i in soup.find_all('img'):
imgdata = i['src']
print(imgdata)
I created a pics folder manually in the folder where the script is running to store the pictures in it. Than i changed your code in the for loop so its appending urls to the imgdata list. The try exceptblock is there because not every url in the list is valid.
import requests
from bs4 import BeautifulSoup
def getdata(url):
r = requests.get(url)
return r.text
htmldata = getdata('https://www.yahoo.com/')
soup = BeautifulSoup(htmldata, 'html.parser')
imgdata = []
for i in soup.find_all('img'):
imgdata.append(i['src']) # made a change here so its appendig to the list
filename = "pics/picture{}.jpg"
for i in range(len(imgdata)):
print(f"img {i+1} / {len(imgdata)+1}")
# try block because not everything in the imgdata list is a valid url
try:
r = requests.get(imgdata[i], stream=True)
with open(filename.format(i), "wb") as f:
f.write(r.content)
except:
print("Url is not an valid")
foo.write('whatever')
foo.close()
Related
Currently I have a script that can only download the HTML of a given page.
Now I want to download all the files of the web page including HTML, CSS, JS and image files (same as we get with a ctrl-s of any website).
My current code is:
import urllib
url = "https://en.wikipedia.org/wiki/Python_%28programming_language%29"
urllib.urlretrieve(url, "t3.html")
I visited many questions but they are all only downloading the HTML.
The following implementation enables you to get the sub-HTML websites. It can be more developed in order to get the other files you need. I sat the depth variable for you to set the maximum sub_websites that you want to parse to.
import urllib2
from BeautifulSoup import *
from urlparse import urljoin
def crawl(pages, depth=None):
indexed_url = [] # a list for the main and sub-HTML websites in the main website
for i in range(depth):
for page in pages:
if page not in indexed_url:
indexed_url.append(page)
try:
c = urllib2.urlopen(page)
except:
print "Could not open %s" % page
continue
soup = BeautifulSoup(c.read())
links = soup('a') #finding all the sub_links
for link in links:
if 'href' in dict(link.attrs):
url = urljoin(page, link['href'])
if url.find("'") != -1:
continue
url = url.split('#')[0]
if url[0:4] == 'http':
indexed_url.append(url)
pages = indexed_url
return indexed_url
pagelist=["https://en.wikipedia.org/wiki/Python_%28programming_language%29"]
urls = crawl(pagelist, depth=2)
print urls
Python3 version, 2019. May this saves some time to somebody:
#!/usr/bin/env python
import urllib.request as urllib2
from bs4 import *
from urllib.parse import urljoin
def crawl(pages, depth=None):
indexed_url = [] # a list for the main and sub-HTML websites in the main website
for i in range(depth):
for page in pages:
if page not in indexed_url:
indexed_url.append(page)
try:
c = urllib2.urlopen(page)
except:
print( "Could not open %s" % page)
continue
soup = BeautifulSoup(c.read())
links = soup('a') #finding all the sub_links
for link in links:
if 'href' in dict(link.attrs):
url = urljoin(page, link['href'])
if url.find("'") != -1:
continue
url = url.split('#')[0]
if url[0:4] == 'http':
indexed_url.append(url)
pages = indexed_url
return indexed_url
pagelist=["https://en.wikipedia.org/wiki/Python_%28programming_language%29"]
urls = crawl(pagelist, depth=1)
print( urls )
You can easily do that with simple python library pywebcopy.
For Current version: 5.0.1
from pywebcopy import save_webpage
url = 'http://some-site.com/some-page.html'
download_folder = '/path/to/downloads/'
kwargs = {'bypass_robots': True, 'project_name': 'recognisable-name'}
save_webpage(url, download_folder, **kwargs)
You will have html, css, js all at your download_folder. Completely working like original site.
Using Python 3+ Requests and other standard libraries.
The function savePage receives a requests.Response and the pagefilename where to save it.
Saves the pagefilename.html on the current folder
Downloads, javascripts, css and images based on the tags script, link and img and saved on a folder pagefilename_files.
Any exception are printed on sys.stderr, returns a BeautifulSoup object .
Requests session must be a global variable unless someone writes a cleaner code here for us.
You can adapt it to your needs.
import os, sys
import requests
from urllib.parse import urljoin
from bs4 import BeautifulSoup
def soupfindAllnSave(pagefolder, url, soup, tag2find='img', inner='src'):
if not os.path.exists(pagefolder): # create only once
os.mkdir(pagefolder)
for res in soup.findAll(tag2find): # images, css, etc..
try:
filename = os.path.basename(res[inner])
fileurl = urljoin(url, res.get(inner))
# rename to saved file path
# res[inner] # may or may not exist
filepath = os.path.join(pagefolder, filename)
res[inner] = os.path.join(os.path.basename(pagefolder), filename)
if not os.path.isfile(filepath): # was not downloaded
with open(filepath, 'wb') as file:
filebin = session.get(fileurl)
file.write(filebin.content)
except Exception as exc:
print(exc, file=sys.stderr)
return soup
def savePage(response, pagefilename='page'):
url = response.url
soup = BeautifulSoup(response.text)
pagefolder = pagefilename+'_files' # page contents
soup = soupfindAllnSave(pagefolder, url, soup, 'img', inner='src')
soup = soupfindAllnSave(pagefolder, url, soup, 'link', inner='href')
soup = soupfindAllnSave(pagefolder, url, soup, 'script', inner='src')
with open(pagefilename+'.html', 'w') as file:
file.write(soup.prettify())
return soup
Example saving google page and its contents (google_files folder)
session = requests.Session()
#... whatever requests config you need here
response = session.get('https://www.google.com')
savePage(response, 'google')
Try the Python library Scrapy. You can program Scrapy to recursively scan a website by downloading its pages, scanning, following links:
An open source and collaborative framework for extracting the data you need from websites. In a fast, simple, yet extensible way.
If I write http://www.chictopia.com/photo/show/3
I can get proper image file.
However, if I set range to crawl image within multiple web page with using for loop
I can't get image file it seems 0bytes file is downloaded
f'http://www.chictopia.com/photo/show/+{x}
why I can get 0bytes image file and could anyone explain how to parse image of multiple page.
Thank you
import re
import requests
from bs4 import BeautifulSoup
for x in range (3,6):
response = requests.get(f'http://www.chictopia.com/photo/show/+{x}')
print (response)
soup = BeautifulSoup(response.text, 'html.parser')
img_tags = soup.find_all('img')
urls = [img['src'] for img in img_tags]
for url in urls:
filename = re.search(r'/([\w_-]+[400]+[.](jpg))$', url)
if not filename:
print("fail".format(url))
continue
with open(filename.group(1), 'wb') as f:
if 'http' not in url:
url = '{}{}'.format(response, url)
response = requests.get(url)
f.write(response.content)
try this
modified the regex pattern and used changed the call to proper image url.
now this code will save all the images containing _400.jpg in their link as following name.
import re
import requests
from bs4 import BeautifulSoup
import shutil
for x in range (3,6):
response = requests.get(f'http://www.chictopia.com/photo/show/+{x}')
# print (response.status_code)
soup = BeautifulSoup(response.text, 'html.parser')
img_tags = soup.find_all('img')
urls = [img['src'] for img in img_tags]
for url in urls:
filename = re.findall(r'(.+_400\.jpg)', url)
if len(filename) != 0:
image = filename[0]
image_name = f"image_{image.split('/')[-1]}"
response = requests.get(image, stream=True)
with open(image_name, 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
print(f'Saved : {image_name}')
for example.
http://images2.chictopia.com/photos/mikajones/2162299642/2162299642_400.jpg -> as image_2162299642_400.jpg
now whats wrong with your code:
you used wrong regex and took the matching value which is not the complete url of the image that's why you are getting the null value for image size (you are not even calling the image url).
all fixed.
Hello Community I have a problem and I dont know how to solve it my problem is I write a script to crawl webpages for Images with BeautifuleSoup4 but I got the error (AttributeError: 'NoneType' object has no attribute 'group')
import re
import requests
from bs4 import BeautifulSoup
site = 'https://www.fotocommunity.de/natur/wolken/3144?sort=new'
response = requests.get(site)
soup = BeautifulSoup(response.text, 'html.parser')
img_tags = soup.find_all('img', {"src": True})
urls = [img["src"] for img in img_tags]
for url in urls:
filename = re.search(r'([\w_-]+[.](jpg|png))$', url)
with open(filename.group(1), 'wb') as f:
if 'http' not in url:
# sometimes an image source can be relative
# if it is provide the base url which also happens
# to be the site variable atm.
url = '{}{}'.format(site, url)
response = requests.get(url)
f.write(response.content)
Your regex is wrong. Use Python's internal urllib to do the heavyweight lifting instead of writing regexes if you're not familiar with them.
Use something like this (untested):
import re
import requests
from bs4 import BeautifulSoup
from urllib.parse import urlsplit # import this additional library
from os.path import basename # import this additional library
site = 'https://www.fotocommunity.de/natur/wolken/3144?sort=new'
response = requests.get(site)
soup = BeautifulSoup(response.text, 'html.parser')
images_div = soup.find(id=re.compile(r"fcx-gallery-\w+")) # focus on the div containing the images
if img_tags: # test if img_tags has any data
img_tags = images_div.find_all('img', {"data-src": True}) # get all the images in that div
urls = [img["data-src"] for img in img_tags] # grab sources from data-source
for url in urls:
filename = basename(urlsplit(url).path) # use this instead of a regex
with open(filename, 'wb') as f: # filename is now a string
if 'http' not in url:
# sometimes an image source can be relative
# if it is provide the base url which also happens
# to be the site variable atm.
url = '{}{}'.format(site, url)
response = requests.get(url)
f.write(response.content)
I am trying to extract all the images from below URL, However, I don't understand the HTTP Error 403: Forbidden, Can it be taken care of during error handling, or simply the URL cant be scraped due to limitations?
from bs4 import BeautifulSoup
from urllib.request import urlopen
import urllib.request
def make_soup(url):
html = urlopen(url).read()
return BeautifulSoup(html)
def get_images(url):
soup = make_soup(url)
#this makes a list of bs4 element tags
images = [img for img in soup.findAll('img')]
print (str(len(images)) + "images found.")
print("downloading to current directory ")
#compile our unicode list of image links
image_links = [each.get('src') for each in images]
for each in image_links:
filename=each.split('/')[-1]
urllib.request.urlretrieve(each,filename)
return image_links
get_images("https://opensignal.com/reports/2019/04/uk/mobile-network-experience")
some sites need you to specify User-Agent header
from bs4 import BeautifulSoup
from urllib.request import Request, urlopen
import urllib.request
def make_soup(url):
site = url
hdr = {'User-Agent': 'Mozilla/5.0'}
req = Request(site, headers=hdr)
page = urlopen(req)
return BeautifulSoup(page)
You can use this function for image scraping. using img tag along not useful nowadays .we can implement something like below, that will fulfill the requirement. It's not relay on any tags so wherever image link is present it will grab it.
def extract_ImageUrl(soup_chunk):
urls_found = []
for tags in soup_chunk.find_all():
attributes = tags.attrs
if str(attributes).__contains__('http'):
for links in attributes.values():
if re.match('http.*\.jpg|png',str(links)):
if len(str(links).split()) <=1:
urls_found.append(links)
else:
link = [i.strip() for i in str(links).split() if re.match('http.*\.jpg|png',str(i))]
urls_found = urls_found + link
print("Found {} image links".format(len(urls_found)))
return urls_found
It's an initial thought, require updates to make it very better.
I'm trying to write a Python script to download images from any website. It is working, but inconsistently. Specifically, find_all("img") is not doing so for the second url. The script is:
# works for http://proof.nationalgeographic.com/2016/02/02/photo-of-the-day-best-of-january-3/
# but not http://www.nationalgeographic.com/photography/proof/2017/05/lake-chad-desertification/
import requests
from PIL import Image
from io import BytesIO
from bs4 import BeautifulSoup
def url_to_image(url, filename):
# get HTTP response, open as bytes, save the image
# http://docs.python-requests.org/en/master/user/quickstart/#binary-response-content
req = requests.get(url)
i = Image.open(BytesIO(req.content))
i.save(filename)
# open page, get HTML request and parse with BeautifulSoup
html = requests.get("http://proof.nationalgeographic.com/2016/02/02/photo-of-the-day-best-of-january-3/")
soup = BeautifulSoup(html.text, "html.parser")
# find all JPEGS in our soup and write their "src" attribute to array
urls = []
for img in soup.find_all("img"):
if img["src"].endswith("jpg"):
print("endswith jpg")
urls.append(str(img["src"]))
print(str(img))
jpeg_no = 00
for url in urls:
url_to_image(url, filename="NatGeoPix/" + str(jpeg_no) + ".jpg")
jpeg_no += 1
The images are rendered with JavaScript on the page that is failing.
First render the page with dryscrape
(If you don't want to use dryscrape see Web-scraping JavaScript page with Python )
e.g.
import requests
from PIL import Image
from io import BytesIO
from bs4 import BeautifulSoup
import dryscrape
def url_to_image(url, filename):
# get HTTP response, open as bytes, save the image
# http://docs.python-requests.org/en/master/user/quickstart/#binary-response-content
req = requests.get(url)
i = Image.open(BytesIO(req.content))
i.save(filename)
# open page, get HTML request and parse with BeautifulSoup
session = dryscrape.Session()
session.visit("http://www.nationalgeographic.com/photography/proof/2017/05/lake-chad-desertification/")
response = session.body()
soup = BeautifulSoup(response, "html.parser")
# find all JPEGS in our soup and write their "src" attribute to array
urls = []
for img in soup.find_all("img"):
if img["src"].endswith("jpg"):
print("endswith jpg")
urls.append(str(img["src"]))
print(str(img))
jpeg_no = 00
for url in urls:
url_to_image(url, filename="NatGeoPix/" + str(jpeg_no) + ".jpg")
jpeg_no += 1
But I would also check that you have an absolute URL not a relative one:
import requests
from PIL import Image
from io import BytesIO
from bs4 import BeautifulSoup
import dryscrape
from urllib.parse import urljoin
def url_to_image(url, filename):
# get HTTP response, open as bytes, save the image
# http://docs.python-requests.org/en/master/user/quickstart/#binary-response-content
req = requests.get(url)
i = Image.open(BytesIO(req.content))
i.save(filename)
# open page, get HTML request and parse with BeautifulSoup
base = "http://www.nationalgeographic.com/photography/proof/2017/05/lake-chad-desertification/"
session = dryscrape.Session()
session.visit(base)
response = session.body()
soup = BeautifulSoup(response, "html.parser")
# find all JPEGS in our soup and write their "src" attribute to array
urls = []
for img in soup.find_all("img"):
if img["src"].endswith("jpg"):
print("endswith jpg")
urls.append(str(img["src"]))
print(str(img))
jpeg_no = 00
for url in urls:
if url.startswith( 'http' ):
absoute = url
else:
absoute = urljoin(base, url)
print (absoute)
url_to_image(absoute, filename="NatGeoPix/" + str(jpeg_no) + ".jpg")
jpeg_no += 1