Unable to save downloaded images into a folder on the desktop using python

Question

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'"

Answer 1

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

Answer 2

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

Answer 3

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()

Answer 4

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.