Currently im facing following problem:
I have 3 download links in a list. Only the last file in the list is downloaded completely.
The others have a file size of one kilobyte.
Code:
from requests import get
def download(url, filename):
with open(filename, "wb") as file:
response = get(url, stream=True)
file.write(response.content)
for link in f:
url = link
split_url = url.split("/")
filename = split_url[-1]
filename = filename.replace("\n", "")
download(url,filename)
The result looks like this:
Result
How do I make sure that all files are downloaded correctly?
All links are direct download links.
Thanks in advance!
EDIT:
I discovered it only happens when I read the links from the .txt
If I create the list in python like this:
links = ["http://ipv4.download.thinkbroadband.com/20MB.zip",
"http://ipv4.download.thinkbroadband.com/10MB.zip",
"http://ipv4.download.thinkbroadband.com/5MB.zip"]
... the problem doesnt appear.
reproduceable example:
from requests import get
def download(url, filename):
with open(filename, "wb") as file:
response = get(url, stream = True)
file.write(response.content)
f = open('links.txt','r')
for link in f:
url = link
split_url = url.split("/")
filename = split_url[-1]
filename = filename.replace("\n", "")
download(url,filename)
content of links.txt:
http://ipv4.download.thinkbroadband.com/20MB.zip
http://ipv4.download.thinkbroadband.com/10MB.zip
http://ipv4.download.thinkbroadband.com/5MB.zip
url = url.replace("\n", "")
solved it!
I'm using the following Python code to download images from a certain website. It's part of a code that I'm using to make a web scraper.
for url in links:
# Invoke wget download method to download specified url image.
local_image_filename = wget.download(url)
# Print out local image file name.
local_image_filename
continue
It's working well, but I want to know if it's possible to add a string as a prefix to each file...
My ideia is get the page title via Xpath and add as a prefix for each file.
I don't know where to add a string in this code. Can someone help me?
For example, I'm downloading these files:
logo.jpg, plans.jpg, circle.jpg
And I need to add a prefix, like these:
Beautiful_Plan_logo.jpg, Beautiful_Plan_plans.jpg, Beautiful_Plan_circle.jpg
Following I'll put the entire code:
import requests
import bs4 as bs
import urllib.request
import wget
##################################################
# getting url images #
##################################################
url = "https://tyreehouseplans.com/shop/house-plans/blackberry-blossom/"
opener = urllib.request.build_opener()
opener.add_headers = [{'User-Agent' : 'Mozilla'}]
urllib.request.install_opener(opener)
raw = requests.get(url).text
soup = bs.BeautifulSoup(raw, 'html.parser')
imgs = soup.find_all('img')
links = []
for img in imgs:
link = img.get('src')
links.append(link)
print(links)
################################################
# downloading images #
################################################
for url in links:
# Invoke wget download method to download specified url image.
local_image_filename = wget.download(url)
# Print out local image file name.
local_image_filename
continue
Thank you for any help!
python module wget has an option out, which determines the name of the output file. For example, the following script downloads 3 images, adding a prefix Beautiful_Plan_.
import wget
base_url = 'https://homepages.cae.wisc.edu/~ece533/images/'
image_names = ['airplane.png', 'arctichare.png', 'baboon.png']
prefix = 'Beautiful_Plan_'
for image_name in image_names:
wget.download(base_url + image_name, out = prefix + image_name)
you can use shutil for this
import shutil
prefix = "prefix_"
#your piece of code
for url in links:
# Invoke wget download method to download specified url image.
local_image_filename = wget.download(url)
# Print out local image file name.
local_image_filename
shutil.copy(local_image_filename, prefix+local_image_filename)
use os.rename as per this documentation
I wrote code for making a seperate file with the extra information up front with a seperator.
import requests
import bs4 as bs
import urllib.request
import wget
##################################################
# getting url images #
##################################################
url = "https://tyreehouseplans.com/shop/house-plans/blackberry-blossom/"
opener = urllib.request.build_opener()
opener.add_headers = [{'User-Agent': 'Mozilla'}]
urllib.request.install_opener(opener)
raw = requests.get(url).text
soup = bs.BeautifulSoup(raw, 'html.parser')
imgs = soup.find_all('img')
links = []
for img in imgs:
link = img.get('src')
links.append(link)
# print(links)
################################################
# downloading images #
################################################
for url in links:
# Invoke wget download method to download specified url image.
try:
local_image_filename = wget.download(url)
except ValueError:
break
# Print out local image file name.
print(local_image_filename)
with open(local_image_filename, 'r') as myFile:
try:
data = myFile.read()
except UnicodeDecodeError:
data = "UNICODE DECODE ERROR"
except ValueError:
data = "VALUE ERROR"
print(data)
print(type(data))
myFile.close()
newSaveString = str(local_image_filename) + "SeperatorOfSomeKind" + str(data)
newFileName = "NEW_" + local_image_filename
with open(newFileName, 'w') as myFile:
myFile.write(newSaveString)
myFile.close()
continue
Firstly really sorry for explaining the problem not clearly in Title. So Let's begin;
I need this captcha image to be downloaded in programmatically way.
import grab, requests, urllib
root_url = 'https://e-okul.meb.gov.tr/'
g = grab.Grab()
g.go(root_url)
e = g.doc.select('//*[#id="image1"]')
captcha_url = root_url + e.attr('src')
img = urllib.request.urlopen(captcha_url)
localFile = open('captcha.jpg', 'wb')
localFile.write(img.read())
localFile.close()
And the result is this.
When I manually download the image with the very known way Save image as..
There is no problem.
Is there any chance to download this captcha with the way that I actually need?
The captcha image depends on a cookie to populate the value that appears on the image.
You should use the same Grab object you loaded the homepage with to also download the captcha image.
Try this:
import grab, requests, urllib
root_url = 'https://e-okul.meb.gov.tr/'
g = grab.Grab()
g.go(root_url)
e = g.doc.select('//*[#id="image1"]')
captcha_url = root_url + e.attr('src')
resp = g.go(captcha_url)
localFile = open('captcha.jpg', 'wb')
localFile.write(resp.body)
localFile.close()
It generated a file with the correct characters in it for me.
More pythonic file writing with:
import grab, requests, urllib
root_url = 'https://e-okul.meb.gov.tr/'
g = grab.Grab()
g.go(root_url)
e = g.doc.select('//*[#id="image1"]')
captcha_url = root_url + e.attr('src')
resp = g.go(captcha_url)
with open('captcha.jpg', 'wb') as localFile
localFile.write(resp.body)
My file named as 'blueberry.jpg' begins downloading, when I click on the following url manually provided that the username and password are typed when asked:
http://example.com/blueberry/download
How can I make that happen using Python?
import urllib.request
url = 'http://example.com/blueberry/download'
data = urllib.request.urlopen(url).read()
fo = open('E:\\quail\\' + url.split('/')[1] + '.jpg', 'w')
print (data, file = fo)
fo.close()
However above program does not write the required file, how can I provide the required username and password?
Use requests, which provides a friendlier interface to the various url libraries in Python:
import os
import requests
from urlparse import urlparse
username = 'foo'
password = 'sekret'
url = 'http://example.com/blueberry/download/somefile.jpg'
filename = os.path.basename(urlparse(url).path)
r = requests.get(url, auth=(username,password))
if r.status_code == 200:
with open(filename, 'wb') as out:
for bits in r.iter_content():
out.write(bits)
UPDATE:
For Python3 get urlparse with: from urllib.parse import urlparse
I'm willing to bet you are using basic auth. So try doing the following:
import urllib.request
url = 'http://username:pwd#example.com/blueberry/download'
data = urllib.request.urlopen(url).read()
fo = open('E:\\quail\\' + url.split('/')[1] + '.jpg', 'w')
print (data, file = fo)
fo.close()
Let me know if this works.
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)