Python WebScraper - object has no attribute 'urlretrieve' - python

I am trying to create a python webscraper that downloads a certain amount of images from a url, to my current directory. However for the following code:
urllib.request.urlretrieve(each, filename)
It is saying that: AttributeError: 'function' object has no attribute 'urlretrieve' when running the program
Here is the full code:
from urllib.request import urlopen
from bs4 import BeautifulSoup as soup
url = 'https://unsplash.com/s/photos/download'
def download_imgs(url, amountOfImgs):
html = urlopen(url).read()
#parsing the html from the url
page_soup = soup(html, "html.parser")
images = [img for img in page_soup.findAll('img')]
counter = 0
#compiling the unicode list of image links
image_links = [each.get('src') for each in images]
for each in image_links:
if(counter <= amountOfImgs):
filename = each.split('/')[-1]
urllib.request.urlretrieve(each, filename)
counter += 1
else:
return image_links
print(download_imgs(url, 5))

It looks like when you imported just URLOpen, you missed everything else.
I did it a bit differently, I got the html using the requests.get method, and removed the need for url open, you could just do
import urlopen, urlretrieve
if you want to use mine, I know it worked,
import urllib.request
from bs4 import BeautifulSoup as soup
import requests
url = 'https://unsplash.com/s/photos/download'
def download_imgs(url, amountOfImgs):
req=requests.get(url)
html=req.text
#parsing the html from the url
page_soup = soup(html, "html.parser")
images = [img for img in page_soup.findAll('img')]
counter = 0
#compiling the unicode list of image links
image_links = [each.get('src') for each in images]
for each in image_links:
if(counter <= amountOfImgs):
filename = each.split('/')[-1]
urllib.request.urlretrieve(each, filename)
counter += 1
else:
return image_links
print(download_imgs(url, 5))

Related

Downloading multiple files from a URL Directory with a similar name

I am trying to download multiple .csv files from an url directory with similar names (AB_daily.csv, BC_daily.csv, etc.). However, each file is stored in different folders in the directory. I know there is a way to use a loop to extract the files, but I can't figure out how to do it with Beautiful Soup or glob. Do you have any suggestions? I've also used pandas.read_csv() to look for shortcuts as I'm just trying to concatenate the files together later. Thank you.
URL Directory: https://dd.weather.gc.ca/hydrometric/csv/
import os
import requests
import urllib.request
from bs4 import BeautifulSoup
def main(url):
with requests.Session() as req:
r = req.get(url)
soup = BeautifulSoup(r.content, 'html.parser')
target = [f"{url[:20]}{item['href']}" for item in soup.select(
"a[href$='AB_daily_hydrometric.csv']")
for x in target:
print(f"Downloading {x}")
r = req.get(x)
name = x.rsplit("/", 1)[-1]
with open(name, 'wb') as f:
f.write(r.content)
main('https://dd.weather.gc.ca/hydrometric/csv')
# For this specific task this will work:
import os
import requests
import urllib.request
from bs4 import BeautifulSoup
csv_links = []
links = ["AB/", "BC/", "MB/", "NB/", "NL/", "NS/", "NT/", "NU/" "ON/", " PE/", "QC/", "SK/", "YT/"]
def main(url):
with requests.Session() as req:
r = req.get(url)
soup = BeautifulSoup(r.content, 'html.parser')
for item in links:
r_daily = BeautifulSoup(req.get(url + item + "daily/").content, 'html.parser')
r_hourly = BeautifulSoup(req.get(url + item + "hourly/").content, 'html.parser')
for item_d in r_daily.find_all('a'):
if ".csv" in item_d.get('href'):
csv_links.append(url + item + "daily/" +item_d.get('href'))
for item_h in r_hourly.find_all('a'):
if ".csv" in item_h.get('href'):
csv_links.append(url + item + "hourly/" + item_h.get('href'))
for x in csv_links:
print(f"Downloading {x}")
r = req.get(x)
name = x.rsplit("/", 1)[-1]
with open(name, 'wb') as f:
f.write(r.content)
main('https://dd.weather.gc.ca/hydrometric/csv/')

File names are cut while download to folder

My current code is cutting first 6 characters from file names while downloading PDF's. So for example PDF file name is 123456acII.pdf (https://example.com/wp-content/uploads/2016/11/123456acII.pdf) but file in folder is acII.pdf.
How to make names be as they are?
import os
import requests
from urllib.parse import urljoin
from bs4 import BeautifulSoup
main = "https://example.com/"
#If there is no such folder, the script will create one automatically
folder_location = r'C:\temp\webscraping'
if not os.path.exists(folder_location):os.mkdir(folder_location)
def Get_Links():
r = requests.get(main).text
soup = BeautifulSoup(r, 'html.parser')
links = []
for item in soup.findAll("div", {'class': 'large-4 medium-4 columns'}):
for n in item.find_all('a'):
print ('Link: '+ n.get('href'))
links.append(n.get('href'))
return links
def Parse_Links():
pdf = set()
for url in Get_Links():
r = requests.get(url).text
soup = BeautifulSoup(r, 'html.parser')
for item in soup.findAll("div", {'class': 'large-6 medium-8 columns large-centered'}):
for link in item.findAll("a"):
link = link.get("href")
if link:
pdf.add(link)
return pdf
def Save():
for item in Parse_Links():
print(f"Downloading File: {item[55:]}")
filename = os.path.join(folder_location,f"{item[55:]}")
r = requests.get(item)
with open(filename, 'wb') as f:
f.write(r.content)
print("done")
Save()
It looks like you are slicing the string starting at index position 55 {item[55:]}. Try to see if it's simply just starting your index position 6 positions prior:
change to: {item[49:]}

Unknown URL Type: Image Scraping

My error:
File "C:/Users/hp dv4/PycharmProjects/project/imagescrap.py", line
22, in
imagefile.write(urllib.request.urlopen(img_src).read())
ValueError: unknown url type: '/img/logo_with_text.png'
I am getting this error while crawling through the specified website whereas, this same code works fine with some other website.
import urllib.request
from bs4 import BeautifulSoup
def make_soup(url):
thepage = urllib.request.urlopen(url)
soupdata = BeautifulSoup(thepage, "html.parser")
return soupdata
i = 1
soup = make_soup("http://ioe.edu.np/")
unique_srcs = []
for img in soup.findAll('img'):
if img.get('src') not in unique_srcs:
unique_srcs.append(img.get('src'))
for img_src in unique_srcs:
filename = str(i)
i = i + 1
imagefile = open(filename + '.png', 'wb')
imagefile.write(urllib.request.urlopen(img_src).read())
imagefile.close()
the above code will encounter one more error.
you are trying to save every file with .png extension, which may make the files unreadable.
import urllib.request
from bs4 import BeautifulSoup
def make_soup(url):
thepage = urllib.request.urlopen(url)
soupdata = BeautifulSoup(thepage, "html.parser")
return soupdata
base_url = "http://ioe.edu.np/"
soup = make_soup(base_url)
unique_srcs = []
for img in soup.findAll('img'):
if img.get('src') not in unique_srcs:
unique_srcs.append(img.get('src'))
for i, img_src in enumerate(unique_srcs):
print(img_src)
filename = str(i)
extension = img_src.split('.')[-1]
with open(filename+'.'+extension, 'wb') as f:
f.write(urllib.request.urlopen(base_url+img_src).read())
few idiomatic python suggestions:
use enumerate instead of trying to manage a counter.
use the with-open construct which takes care of closing your file.
one other thing you could do to further improve:
use a set instead of a list, so that you don't download the same file twice.
As the error messages says:
unknown url type: '/img/logo_with_text.png'
add http://ioe.edu.np/ in front of img_src and it should work

How can I improve downloading speed with python urllib.request

How can I improve downloading speed with urllib.request? I want to download image from web and It works well. But it takes too long downloading it. It took 42 seconds to excute donwload_album_art() func. What Can I do for that? Can I use mutiprocess or etc? h
import os
import shutil
import requests
from bs4 import BeautifulSoup
from urllib import request
URL = 'https://music.bugs.co.kr/chart/track/day/total'
PATH = os.getcwd() + '/static/images/'
# Scrapping html code
def get_html(target_url):
_html = ""
response = requests.get(target_url)
if response.status_code == 200:
_html = response.text
return _html
# parse image url and save in list
def get_image_url():
html = get_html(URL)
soup = BeautifulSoup(html, 'html.parser')
img_url = []
for image in soup.select('a.thumbnail > img'):
if image.has_attr('src'):
img_url.append(image.get('src'))
else:
continue
return img_url
# download album art in static/images directory
def download_album_arts():
images = get_image_url()
for i in range(0, 100):
url = images[i]
file_name = PATH + str(i + 1) + '.png'
request.urlretrieve(url, file_name)
# delete all album art
def delete_album_art():
path = os.getcwd() + '/static/images'
if os.path.exists(path):
shutil.rmtree(path)
os.mkdir(path)
else:
os.mkdir(path)

Save image from url to special folder

I want to save images from url to special folder, for example 'my_images', but not to default(where my *.py file is). Is it possible to make it?
Because my code saves all images to folder with *.py file.
Here is my code:
import urllib.request
from bs4 import BeautifulSoup
import re
import os
BASE_URL = 'https://fachowiec.com/sklep/pl/products/index?Products_page=1&pageSize=15'
def get_domain(url):
domain = re.findall(r'https:\W\W\w+\.\w+', url)
return domain[0]
def get_html(url):
request = urllib.request.urlopen(url)
return request.read()
def get_img(html):
soup = BeautifulSoup(html)
img_box = []
imgs = soup.find_all('div', class_= 'pthumb')
for img in imgs:
img_box.append(get_domain(BASE_URL) + img.img['src'])
for img in img_box:
urllib.request.urlretrieve(img, os.path.basename(img))
def main():
get_img(get_html('https://fachowiec.com/sklep/pl/products/index?Products_page=1&pageSize=15'))
if __name__ == '__main__':
main()
def get_img(html):
soup = BeautifulSoup(html)
img_box = []
imgs = soup.find_all('div', class_= 'pthumb')
for img in imgs:
img_box.append(get_domain(BASE_URL) + img.img['src'])
my_path = '/home/<username>/Desktop' # use whatever path you like
for img in img_box:
urllib.request.urlretrieve(img, os.path.join(my_path, os.path.basename(img)))
You should add the pathname in second parameter of urllib.request.urlretrieve. Something like below:
urllib.request.urlretrieve(img, "PATH"+os.path.basename(img))
The second argument, if present, specifies the file location to copy to (if absent, the location will be a tempfile with a generated name).

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