I am still new to coding. I needed to write a code to iterate through a data folder of many html files and execute a predefined function (Extracting specific tables from HTML document). I used bs4 to parse the html file. The proposed solution below allowed to retrieve the files and extract the tables from each html file.
from bs4 import BeautifulSoup
import glob
def get_soup(html_file_path):
f = html_file_path.open()
return BeautifulSoup(f, "lxml")
def get_all_tables(soup):
return soup.find_all("table")
def get_all_html_files(root_path):
return Path(root_path).glob("**/*.html")
if __name__ == "__main__":
html_root = Path("data_file_pathname/")
soup = get_soup(html_file)
tables = get_all_tables(soup)
print(f"[+] Found a total of {len(tables)} tables.")
Thanks
You can use the Path.glob function from the pathlib standard library module.
For example:
from pathlib import Path
def get_soup(html_file_path): # added argument
f = html_file_path.open()
return BeautifulSoup(f, "lxml")
def get_all_tables(soup):
return soup.find_all("table")
def get_all_html_files(root_path):
return Path(root_path).glob("**/*.html")
if __name__ == "__main__":
html_root = Path("./html_files/") # This is the folder with the html files
for html_file in get_all_html_files(html_root):
soup = get_soup(html_file)
tables = get_all_tables(soup)
I have made a scraper which is at this moment parsing image links and saving downloaded images into python directory by default. The only thing i wanna do now is choose a folder on the desktop to save those images within but can't. Here is what I'm up to:
import requests
import os.path
import urllib.request
from lxml import html
def Startpoint():
url = "https://www.aliexpress.com/"
response = requests.get(url)
tree = html.fromstring(response.text)
titles = tree.xpath('//div[#class="item-inner"]')
for title in titles:
Pics="https:" + title.xpath('.//span[#class="pic"]//img/#src')[0]
endpoint(Pics)
def endpoint(images):
sdir = (r'C:\Users\ar\Desktop\mth')
testfile = urllib.request.URLopener()
xx = testfile.retrieve(images, images.split('/')[-1])
filename=os.path.join(sdir,xx)
print(filename)
Startpoint()
Upon execution the above code throws an error showing: "join() argument must be str or bytes, not 'tuple'"
you can download images with urllib of python. You can see the official documentation of python here urllib documentation for python 2.7 . If you want to use python 3 then follow this documentation urllib for python 3
You could use urllib.request, BytesIO from io and PIL Image.
(if you have a direct url to the image)
from PIL import Image
from io import BytesIO
import urllib.request
def download_image(url):
req = urllib.request.Request(url)
response = urllib.request.urlopen(req)
content = response.read()
img = Image.open(BytesIO(content))
img.filename = url
return img
The images are dynamic now. So, I thought to update this post:
import os
from selenium import webdriver
import urllib.request
from lxml.html import fromstring
url = "https://www.aliexpress.com/"
def get_data(link):
driver.get(link)
tree = fromstring(driver.page_source)
for title in tree.xpath('//li[#class="item"]'):
pics = "https:" + title.xpath('.//*[contains(#class,"img-wrapper")]//img/#src')[0]
os.chdir(r"C:\Users\WCS\Desktop\test")
urllib.request.urlretrieve(pics, pics.split('/')[-1])
if __name__ == '__main__':
driver = webdriver.Chrome()
get_data(url)
driver.quit()
This is the code to download the html file from the web
import random
import urllib.request
def download(url):
name = random.randrange(1, 1000)
#this is the random function to give the name to the file
full_name = str(name) + ".html" #compatible data type
urllib.request.urlretrieve(url,full_name) #main function
download("any url")
This is the code for downloading any html file from the internet just you have to provide the link in the function.
As in your case you have told that you have retrieved the images links from the web page So you can change the extension from ".html" to compatible type, but the problem is that the image can be of different extension may be ".jpg" , ".png" etc.
So what you can do is you can match the ending of the link using if else with string matching and then assign the extension in the end.
Here is the example for the illustration
import random
import urllib.request
if(link extension is ".png"): #pseudo code
def download(url):
name = random.randrange(1, 1000)
#this is the random function to give the name to the file
full_name = str(name) + ".png" #compatible extension with .png
urllib.request.urlretrieve(url,full_name) #main function
download("any url")
else if (link extension is ".jpg"): #pseudo code
def download(url):
name = random.randrange(1, 1000)
#this is the random function to give the name to the file
full_name = str(name) + ".jpg" #compatible extension with .jpg
urllib.request.urlretrieve(url,full_name) #main function
download("any url")
You can use multiple if else for the various type of the extension.
If it helps for your situation have a Thumbs up buddy.
the task is easy: use Python to download all PDFs from:
https://www.electroimpact.com/Company/Patents.aspx
I am just a beginner of Python. I read python crawler but samples deal with html not aspx. And all I got is blank file downloaded.
Following is my code:
import urllib2
import re
def saveFile(url, fileName):
request = urllib2.Request(url)
response = urllib2.urlopen(request)
with open(fileName,'wb') as handle:
handle.write(response.read())
def main():
base_url = 'https://www.electroimpact.com/Company/Patents/'
page = 'https://www.electroimpact.com/Company/Patents.aspx'
request = urllib2.Request(page)
response = urllib2.urlopen(request)
url_lst = re.findall('href.*(US.*\.pdf)', response.read())
print url_lst
Result:
['US5201205.pdf', 'US5279024.pdf', 'US5339598.pdf', 'US9021688B2.pdf']
Only 4 PDFs were found by my regular expression. Actually, there are much more PDFs to extract. Why?
With lxml.html and cssselect instead of re you will get all linked patent document paths:
#!/usr/bin/env python
# coding: utf8
from __future__ import absolute_import, division, print_function
import urllib2
from lxml import html
def main():
url = 'https://www.electroimpact.com/Company/Patents.aspx'
source = urllib2.urlopen(url).read()
document = html.fromstring(source)
patent_paths = [
a.attrib['href'] for a in document.cssselect('div.PatentNumber a')
]
print(patent_paths)
if __name__ == '__main__':
main()
I need to store HTML file as text files. In the same name of web tittle. something went wrong with my code so that it is not creating file in side the directory. I have directory permission to write. I am using Ubuntu 12.04LTS
Directory /home/user1/
File name
print name = Mathrubhumi Sports - ശ്രീനിക്ക് പച്ചക്കൊടി
The file name contain Unicode values
import os
from urllib import urlopen
from bs4 import BeautifulSoup
url= "http://www.mathrubhumi.com/sports/story.php?id=397111"
raw = urlopen(url).read()
soup = BeautifulSoup(raw,'lxml')
texts = soup.findAll(text=True)
name = soup.title.text
name= name+'.txt'
def contains_unicode(text):
try:
str(text)
except:
return True
return False
result = ''.join((text for text in texts if contains_unicode(text)))
# Output to a file
with open(os.path.join('/home/user1/textfiles',name,'w') as out:
out.write(result)
Please help me to debug it
I tried this and it worked, it created a file called Mathrub*.txt with some text in it in the current directory.
import codecs
import os
from urllib import urlopen
from bs4 import BeautifulSoup
url= "http://www.mathrubhumi.com/sports/story.php?id=397111"
raw = urlopen(url).read()
soup = BeautifulSoup(raw,'lxml')
texts = soup.findAll(text=True)
name = soup.title.string
name= name+'.txt'
def contains_unicode(text):
try:
str(text)
except:
return True
return False
result = ''.join((text for text in texts if contains_unicode(text)))
# Output to a file
with codecs.open(name,'w',encoding="utf-8") as out:
out.write(result)
Before adding the codecs part, it complained loudly about trying to write some characters that it did not know how to interpret.
I know the URL of an image on Internet.
e.g. http://www.digimouth.com/news/media/2011/09/google-logo.jpg, which contains the logo of Google.
Now, how can I download this image using Python without actually opening the URL in a browser and saving the file manually.
Python 2
Here is a more straightforward way if all you want to do is save it as a file:
import urllib
urllib.urlretrieve("http://www.digimouth.com/news/media/2011/09/google-logo.jpg", "local-filename.jpg")
The second argument is the local path where the file should be saved.
Python 3
As SergO suggested the code below should work with Python 3.
import urllib.request
urllib.request.urlretrieve("http://www.digimouth.com/news/media/2011/09/google-logo.jpg", "local-filename.jpg")
import urllib
resource = urllib.urlopen("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")
output = open("file01.jpg","wb")
output.write(resource.read())
output.close()
file01.jpg will contain your image.
I wrote a script that does just this, and it is available on my github for your use.
I utilized BeautifulSoup to allow me to parse any website for images. If you will be doing much web scraping (or intend to use my tool) I suggest you sudo pip install BeautifulSoup. Information on BeautifulSoup is available here.
For convenience here is my code:
from bs4 import BeautifulSoup
from urllib2 import urlopen
import urllib
# use this image scraper from the location that
#you want to save scraped images to
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 images to current working 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.urlretrieve(each, filename)
return image_links
#a standard call looks like this
#get_images('http://www.wookmark.com')
This can be done with requests. Load the page and dump the binary content to a file.
import os
import requests
url = 'https://apod.nasa.gov/apod/image/1701/potw1636aN159_HST_2048.jpg'
page = requests.get(url)
f_ext = os.path.splitext(url)[-1]
f_name = 'img{}'.format(f_ext)
with open(f_name, 'wb') as f:
f.write(page.content)
Python 3
urllib.request — Extensible library for opening URLs
from urllib.error import HTTPError
from urllib.request import urlretrieve
try:
urlretrieve(image_url, image_local_path)
except FileNotFoundError as err:
print(err) # something wrong with local path
except HTTPError as err:
print(err) # something wrong with url
I made a script expanding on Yup.'s script. I fixed some things. It will now bypass 403:Forbidden problems. It wont crash when an image fails to be retrieved. It tries to avoid corrupted previews. It gets the right absolute urls. It gives out more information. It can be run with an argument from the command line.
# getem.py
# python2 script to download all images in a given url
# use: python getem.py http://url.where.images.are
from bs4 import BeautifulSoup
import urllib2
import shutil
import requests
from urlparse import urljoin
import sys
import time
def make_soup(url):
req = urllib2.Request(url, headers={'User-Agent' : "Magic Browser"})
html = urllib2.urlopen(req)
return BeautifulSoup(html, 'html.parser')
def get_images(url):
soup = make_soup(url)
images = [img for img in soup.findAll('img')]
print (str(len(images)) + " images found.")
print 'Downloading images to current working directory.'
image_links = [each.get('src') for each in images]
for each in image_links:
try:
filename = each.strip().split('/')[-1].strip()
src = urljoin(url, each)
print 'Getting: ' + filename
response = requests.get(src, stream=True)
# delay to avoid corrupted previews
time.sleep(1)
with open(filename, 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
except:
print ' An error occured. Continuing.'
print 'Done.'
if __name__ == '__main__':
url = sys.argv[1]
get_images(url)
A solution which works with Python 2 and Python 3:
try:
from urllib.request import urlretrieve # Python 3
except ImportError:
from urllib import urlretrieve # Python 2
url = "http://www.digimouth.com/news/media/2011/09/google-logo.jpg"
urlretrieve(url, "local-filename.jpg")
or, if the additional requirement of requests is acceptable and if it is a http(s) URL:
def load_requests(source_url, sink_path):
"""
Load a file from an URL (e.g. http).
Parameters
----------
source_url : str
Where to load the file from.
sink_path : str
Where the loaded file is stored.
"""
import requests
r = requests.get(source_url, stream=True)
if r.status_code == 200:
with open(sink_path, 'wb') as f:
for chunk in r:
f.write(chunk)
Using requests library
import requests
import shutil,os
headers = {
'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/78.0.3904.108 Safari/537.36'
}
currentDir = os.getcwd()
path = os.path.join(currentDir,'Images')#saving images to Images folder
def ImageDl(url):
attempts = 0
while attempts < 5:#retry 5 times
try:
filename = url.split('/')[-1]
r = requests.get(url,headers=headers,stream=True,timeout=5)
if r.status_code == 200:
with open(os.path.join(path,filename),'wb') as f:
r.raw.decode_content = True
shutil.copyfileobj(r.raw,f)
print(filename)
break
except Exception as e:
attempts+=1
print(e)
ImageDl(url)
Use a simple python wget module to download the link. Usage below:
import wget
wget.download('http://www.digimouth.com/news/media/2011/09/google-logo.jpg')
This is very short answer.
import urllib
urllib.urlretrieve("http://photogallery.sandesh.com/Picture.aspx?AlubumId=422040", "Abc.jpg")
Version for Python 3
I adjusted the code of #madprops for Python 3
# getem.py
# python2 script to download all images in a given url
# use: python getem.py http://url.where.images.are
from bs4 import BeautifulSoup
import urllib.request
import shutil
import requests
from urllib.parse import urljoin
import sys
import time
def make_soup(url):
req = urllib.request.Request(url, headers={'User-Agent' : "Magic Browser"})
html = urllib.request.urlopen(req)
return BeautifulSoup(html, 'html.parser')
def get_images(url):
soup = make_soup(url)
images = [img for img in soup.findAll('img')]
print (str(len(images)) + " images found.")
print('Downloading images to current working directory.')
image_links = [each.get('src') for each in images]
for each in image_links:
try:
filename = each.strip().split('/')[-1].strip()
src = urljoin(url, each)
print('Getting: ' + filename)
response = requests.get(src, stream=True)
# delay to avoid corrupted previews
time.sleep(1)
with open(filename, 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
except:
print(' An error occured. Continuing.')
print('Done.')
if __name__ == '__main__':
get_images('http://www.wookmark.com')
Late answer, but for python>=3.6 you can use dload, i.e.:
import dload
dload.save("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")
if you need the image as bytes, use:
img_bytes = dload.bytes("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")
install using pip3 install dload
Something fresh for Python 3 using Requests:
Comments in the code. Ready to use function.
import requests
from os import path
def get_image(image_url):
"""
Get image based on url.
:return: Image name if everything OK, False otherwise
"""
image_name = path.split(image_url)[1]
try:
image = requests.get(image_url)
except OSError: # Little too wide, but work OK, no additional imports needed. Catch all conection problems
return False
if image.status_code == 200: # we could have retrieved error page
base_dir = path.join(path.dirname(path.realpath(__file__)), "images") # Use your own path or "" to use current working directory. Folder must exist.
with open(path.join(base_dir, image_name), "wb") as f:
f.write(image.content)
return image_name
get_image("https://apod.nasddfda.gov/apod/image/2003/S106_Mishra_1947.jpg")
this is the easiest method to download images.
import requests
from slugify import slugify
img_url = 'https://apod.nasa.gov/apod/image/1701/potw1636aN159_HST_2048.jpg'
img = requests.get(img_url).content
img_file = open(slugify(img_url) + '.' + str(img_url).split('.')[-1], 'wb')
img_file.write(img)
img_file.close()
If you don't already have the url for the image, you could scrape it with gazpacho:
from gazpacho import Soup
base_url = "http://books.toscrape.com"
soup = Soup.get(base_url)
links = [img.attrs["src"] for img in soup.find("img")]
And then download the asset with urllib as mentioned:
from pathlib import Path
from urllib.request import urlretrieve as download
directory = "images"
Path(directory).mkdir(exist_ok=True)
link = links[0]
name = link.split("/")[-1]
download(f"{base_url}/{link}", f"{directory}/{name}")
# import the required libraries from Python
import pathlib,urllib.request
# Using pathlib, specify where the image is to be saved
downloads_path = str(pathlib.Path.home() / "Downloads")
# Form a full image path by joining the path to the
# images' new name
picture_path = os.path.join(downloads_path, "new-image.png")
# "/home/User/Downloads/new-image.png"
# Using "urlretrieve()" from urllib.request save the image
urllib.request.urlretrieve("//example.com/image.png", picture_path)
# urlretrieve() takes in 2 arguments
# 1. The URL of the image to be downloaded
# 2. The image new name after download. By default, the image is saved
# inside your current working directory
Ok, so, this is my rudimentary attempt, and probably total overkill.
Update if needed, as this doesn't handle any timeouts, but, I got this working for fun.
Code listed here: https://github.com/JayRizzo/JayRizzoTools/blob/master/pyImageDownloader.py
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# =============================================================================
# Created Syst: MAC OSX High Sierra 21.5.0 (17G65)
# Created Plat: Python 3.9.5 ('v3.9.5:0a7dcbdb13', 'May 3 2021 13:17:02')
# Created By : Jeromie Kirchoff
# Created Date: Thu Jun 15 23:31:01 2022 CDT
# Last ModDate: Thu Jun 16 01:41:01 2022 CDT
# =============================================================================
# NOTE: Doesn't work on SVG images at this time.
# I will look into this further: https://stackoverflow.com/a/6599172/1896134
# =============================================================================
import requests # to get image from the web
import shutil # to save it locally
import os # needed
from os.path import exists as filepathexist # check if file paths exist
from os.path import join # joins path for different os
from os.path import expanduser # expands current home
from pyuser_agent import UA # generates random UserAgent
class ImageDownloader(object):
"""URL ImageDownloader.
Input : Full Image URL
Output: Image saved to your ~/Pictures/JayRizzoDL folder.
"""
def __init__(self, URL: str):
self.url = URL
self.headers = {"User-Agent" : UA().random}
self.currentHome = expanduser('~')
self.desktop = join(self.currentHome + "/Desktop/")
self.download = join(self.currentHome + "/Downloads/")
self.pictures = join(self.currentHome + "/Pictures/JayRizzoDL/")
self.outfile = ""
self.filename = ""
self.response = ""
self.rawstream = ""
self.createdfilepath = ""
self.imgFileName = ""
# Check if the JayRizzoDL exists in the pictures folder.
# if it doesn't exist create it.
if not filepathexist(self.pictures):
os.mkdir(self.pictures)
self.main()
def getFileNameFromURL(self, URL: str):
"""Parse the URL for the name after the last forward slash."""
NewFileName = self.url.strip().split('/')[-1].strip()
return NewFileName
def getResponse(self, URL: str):
"""Try streaming the URL for the raw data."""
self.response = requests.get(self.url, headers=self.headers, stream=True)
return self.response
def gocreateFile(self, name: str, response):
"""Try creating the file with the raw data in a custom folder."""
self.outfile = join(self.pictures, name)
with open(self.outfile, 'wb') as outFilePath:
shutil.copyfileobj(response.raw, outFilePath)
return self.outfile
def main(self):
"""Combine Everything and use in for loops."""
self.filename = self.getFileNameFromURL(self.url)
self.rawstream = self.getResponse(self.url)
self.createdfilepath = self.gocreateFile(self.filename, self.rawstream)
print(f"File was created: {self.createdfilepath}")
return
if __name__ == '__main__':
# Example when calling the file directly.
ImageDownloader("https://stackoverflow.design/assets/img/logos/so/logo-stackoverflow.png")
Download Image file, with avoiding all possible error:
import requests
import validators
from urllib.request import Request, urlopen
from urllib.error import URLError, HTTPError
def is_downloadable(url):
valid=validators. url(url)
if valid==False:
return False
req = Request(url)
try:
response = urlopen(req)
except HTTPError as e:
return False
except URLError as e:
return False
else:
return True
for i in range(len(File_data)): #File data Contain list of address for image
#file
url = File_data[i][1]
try:
if (is_downloadable(url)):
try:
r = requests.get(url, allow_redirects=True)
if url.find('/'):
fname = url.rsplit('/', 1)[1]
fname = pth+File_data[i][0]+"$"+fname #Destination to save
#image file
open(fname, 'wb').write(r.content)
except Exception as e:
print(e)
except Exception as e:
print(e)