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Python Requests-html not return the page content

I'm new to Python and would like your advice for the issue I've encountered recently. I'm doing a small project where I tried to scrape a comic website to download a chapter (pictures). However, when printing out the page content for testing (because i tried to use Beautifulsoup.select() and got no result), it only showed a line of html:
'document.cookie="VinaHost-Shield=a7a00919549a80aa44d5e1df8a26ae20"+"; path=/";window.location.reload(true);'
Any help would be really appreciated.
from requests_html import HTMLSession
session = HTMLSession()
res = session.get("https://truyenqqpro.com/truyen-tranh/dao-hai-tac-128-chap-1060.html")
res.html.render()
print(res.content)
I also tried this but the resutl was the same.
import requests, bs4
url = "https://truyenqqpro.com/truyen-tranh/dao-hai-tac-128-chap-1060.html"
res = requests.get(url, headers={"User-Agent": "Requests"})
res.raise_for_status()
# soup = bs4.BeautifulSoup(res.text, "html.parser")
# onePiece = soup.select(".page-chapter")
print(res.content)
update: I installed docker and splash (on Windows 11) and it worked. I included the update code. Thanks Franz and others for yor help.
import os
import requests, bs4
os.makedirs("OnePiece", exist_ok=True)
url = "https://truyenqqpro.com/truyen-tranh/dao-hai-tac-128-chap-1060.html"
res = requests.get("http://localhost:8050/render.html", params={"url": url, "wait": 5})
res.raise_for_status()
soup = bs4.BeautifulSoup(res.text, "html.parser")
onePiece = soup.find_all("img", class_="lazy")
for element in onePiece:
imageLink = "https:" + element["data-cdn"]
res = requests.get(imageLink)
imageFile = open(os.path.join("OnePiece", os.path.basename(imageLink)), "wb")
for chunk in res.iter_content(100000):
imageFile.write(chunk)
imageFile.close()

import urllib.request
request_url = urllib.request.urlopen('https://truyenqqpro.com/truyen-tranh/dao-hai-tac-128-chap-1060.html')
print(request_url.read())
it will return html code of the page.
by the way in that html it is loading several images. you need to use regx to trakdown those img urls and download them.

This response means that we need a javascript render that reload the page using this cookie. for you get the content some workaround must be added.
I commonly use splash scrapinhub render engine and putting a sleep in the page just renders ok all the content. Some tools that render in same way are selenium for python or pupitter in JS.
Link for Splash and Pupeteer

API - Web Scrape

how to get access to this API:
import requests
url = 'https://b2c-api-premiumlabel-production.azurewebsites.net/api/b2c/page/menu?id_loja=2691'
print(requests.get(url))
I'm trying to retrieve data from this site via API, I found the url above and I can see its data , however I can't seem to get it right because I'm running into code 403.
This is the website url:
https://www.nagumo.com.br/osasco-lj46-osasco-ayrosa-rua-avestruz/departamentos
I'm trying to retrieve items category, they are visible for me, but I'm unable to take them.
Later I'll use these categories to iterate over products API.
API Category
Obs: please be gentle it's my first post here =]

To get the data as you shown in your image the following headers and endpoint are needed:
import requests
headers = {
'sm-token': '{"IdLoja":2691,"IdRede":884}',
'User-Agent': 'Mozilla/5.0',
'Referer': 'https://www.nagumo.com.br/osasco-lj46-osasco-ayrosa-rua-avestruz/departamentos',
}
params = {
'id_loja': '2691',
}
r = requests.get('https://www.nagumo.com.br/api/b2c/page/menu', params=params, headers=headers)
r.json()

Not sure exactly what your issue is here.
Bu if you want to see the content of the response and not just the 200/400 reponses. You need to add '.content' to your print.
Eg.
#Create Session
s = requests.Session()
#Example Connection Variables, probably not required for your use case.
setCookieUrl = 'https://www...'
HeadersJson = {'Accept-Language':'en-us'}
bodyJson = {"__type":"xxx","applicationName":"xxx","userID":"User01","password":"password2021"}
#Get Request
p = s.get(otherUrl, json=otherBodyJson, headers=otherHeadersJson)
print(p) #Print response (200 etc)
#print(p.headers)
#print(p.content) #Print the content of the response.
#print(s.cookies)

I'm also new here haha, but besides this requests library, you'll also need another one like beautiful soup for what you're trying to do.
bs4 installation: https:https://www.crummy.com/software/BeautifulSoup/bs4/doc/#installing-beautiful-soup
Once you install it and import it, it's just continuing what you were doing to actively get your data.
response = requests.get(url)
soup = BeautifulSoup(response.text, "html.parser")
this gets the entire HTML content of the page, and so, you can get your data from this page based on their css selectors like this:
site_data = soup.select('selector')
site_data is an array of things with that 'selector', so a simple for loop and an array to add your items in would suffice (as an example, getting links for each book on a bookstore site)
For example, if i was trying to get links from a site:
import requests
from bs4 import BeautifulSoup
sites = []
URL = 'https://b2c-api-premiumlabel-production.azurewebsites.net/api/b2c/page/menu?id_loja=2691'
response = requests.get(url)
soup = BeautifulSoup(response.text, "html.parser")
links = soup.select("a") # list of all items with this selector
for link in links:
sites.append(link)
Also, a helpful tip is when you inspect the page (right click and at the bottom press 'inspect'), you can see the code for the page. Go to the HTML and find the data you want and right click it and select copy -> copy selector. This will make it really easy for you to get the data you want on that site.
helpful sites:
https://oxylabs.io/blog/python-web-scraping
https://realpython.com/beautiful-soup-web-scraper-python/

How to download in python big media links of a web page behind a log in form?

I'm looking for some library or libraries in Python to:
a) log in a web site,
b) find all links to some media files (let us say having "download" in their URLs), and
c) download each file efficiently directly to the hard drive (without loading the whole media file into RAM).
Thanks

You can use the broadly used requests module (more than 35k stars on github), and BeautifulSoup. The former handles session cookies, redirections, encodings, compression and more transparently. The later finds parts in the HTML code and has an easy-to-remember syntax, e.g. [] for properties of HTML tags.
It follows a complete example in Python 3.5.2 for a web site that you can scrap without a JavaScript engine (otherwise you can use Selenium), and downloading sequentially some links with download in its URL.
import shutil
import sys
import requests
from bs4 import BeautifulSoup
""" Requirements: beautifulsoup4, requests """
SCHEMA_DOMAIN = 'https://exmaple.com'
URL = SCHEMA_DOMAIN + '/house.php/' # this is the log-in URL
# here are the name property of the input fields in the log-in form.
KEYS = ['login[_csrf_token]',
'login[login]',
'login[password]']
client = requests.session()
request = client.get(URL)
soup = BeautifulSoup(request.text, features="html.parser")
data = {KEYS[0]: soup.find('input', dict(name=KEYS[0]))['value'],
KEYS[1]: 'my_username',
KEYS[2]: 'my_password'}
# The first argument here is the URL of the action property of the log-in form
request = client.post(SCHEMA_DOMAIN + '/house.php/user/login',
data=data,
headers=dict(Referer=URL))
soup = BeautifulSoup(request.text, features="html.parser")
generator = ((tag['href'], tag.string)
for tag in soup.find_all('a')
if 'download' in tag['href'])
for url, name in generator:
with client.get(SCHEMA_DOMAIN + url, stream=True) as request:
if request.status_code == 200:
with open(name, 'wb') as output:
request.raw.decode_content = True
shutil.copyfileobj(request.raw, output)
else:
print('status code was {} for {}'.format(request.status_code,
name),
file=sys.stderr)

You can use the mechanize module to log into websites like so:
import mechanize
br = mechanize.Browser()
br.set_handle_robots(False)
br.open("http://www.example.com")
br.select_form(nr=0) #Pass parameters to uniquely identify login form if needed
br['username'] = '...'
br['password'] = '...'
result = br.submit().read()
Use bs4 to parse this response and find all the hyperlinks in the page like so:
from bs4 import BeautifulSoup
import re
soup = BeautifulSoup(result, "lxml")
links = []
for link in soup.findAll('a'):
links.append(link.get('href'))
You can use re to further narrow down the links you need from all the links present in the response webpage, which are media links (.mp3, .mp4, .jpg, etc) in your case.
Finally, use requests module to stream the media files so that they don't take up too much memory like so:
response = requests.get(url, stream=True) #URL here is the media URL
handle = open(target_path, "wb")
for chunk in response.iter_content(chunk_size=512):
if chunk: # filter out keep-alive new chunks
handle.write(chunk)
handle.close()
when the stream attribute of get() is set to True, the content does not immediately start downloading to RAM, instead the response behaves like an iterable, which you can iterate over in chunks of size chunk_size in the loop right after the get() statement. Before moving on to the next chunk, you can write the previous chunk to memory hence ensuring that the data isn't stored in RAM.
You will have to put this last chunk of code in a loop if you want to download media of every link in the links list.
You will probably have to end up making some changes to this code to make it work as I haven't tested it for your use case myself, but hopefully this gives a blueprint to work off of.

How to check if (https://) is a image or web link

How to check if a hyperlink is a image link or web link.
image_list = []
url = 'http://www.image.jpg/'
if any(x in '.jpg .gif .png .jpeg' for x in url):
image_list.append(url)
else:
r = requests.get(url)
soup = BeautifulSoup(r.content, "html5lib")
for link in soup.find_all('img'):
src = link.get('src')
if src.startswith("https"):
image_list.append(src)
The code above works in finding out the hyperlink contains image formats, however whenever i use a link that does not contain ".jpg ect..." it still appends the link to the image_list and skips the else statement.

Let's look at this code:
any(x in '.jpg .gif .png .jpeg' for x in url):
This checks if any letter in the URL is in the string. The 'p' from http is in the string, so you will always get a true result.
Here's how you could check the extension of a URL:
import posixpath
import urllib.parse
IMAGE_EXTS = { '.png', '.jpg', '.jpeg', '.gif' }
url = 'http://example.com/'
if posixpath.splitext(urllib.parse.urlparse(url).path)[1] in IMAGE_EXTS:
# Has image extension...
But that's a moot point, because the extension of a URL doesn't tell you whether it's an image. Unlike regular files, for URLs, the extension is completely irrelevant! You can have an .html URL which gives you a PNG image, or a .gif URL which is really an HTML web page. You need to check the Content-Type of the HTTP reply.

Python to list HTTP-files and directories

How can I list files and folders if I only have an IP-address?
With urllib and others, I am only able to display the content of the index.html file. But what if I want to see which files are in the root as well?
I am looking for an example that shows how to implement username and password if needed. (Most of the time index.html is public, but sometimes the other files are not).

Use requests to get page content and BeautifulSoup to parse the result.
For example if we search for all iso files at http://cdimage.debian.org/debian-cd/8.2.0-live/i386/iso-hybrid/:
from bs4 import BeautifulSoup
import requests
url = 'http://cdimage.debian.org/debian-cd/8.2.0-live/i386/iso-hybrid/'
ext = 'iso'
def listFD(url, ext=''):
page = requests.get(url).text
print page
soup = BeautifulSoup(page, 'html.parser')
return [url + '/' + node.get('href') for node in soup.find_all('a') if node.get('href').endswith(ext)]
for file in listFD(url, ext):
print file

You cannot get the directory listing directly via HTTP, as another answer says. It's the HTTP server that "decides" what to give you. Some will give you an HTML page displaying links to all the files inside a "directory", some will give you some page (index.html), and some will not even interpret the "directory" as one.
For example, you might have a link to "http://localhost/user-login/": This does not mean that there is a directory called user-login in the document root of the server. The server interprets that as a "link" to some page.
Now, to achieve what you want, you either have to use something other than HTTP (an FTP server on the "ip address" you want to access would do the job), or set up an HTTP server on that machine that provides for each path (http://192.168.2.100/directory) a list of files in it (in whatever format) and parse that through Python.
If the server provides an "index of /bla/bla" kind of page (like Apache server do, directory listings), you could parse the HTML output to find out the names of files and directories. If not (e.g. a custom index.html, or whatever the server decides to give you), then you're out of luck :(, you can't do it.

Zety provides a nice compact solution. I would add to his example by making the requests component more robust and functional:
import requests
from bs4 import BeautifulSoup
def get_url_paths(url, ext='', params={}):
response = requests.get(url, params=params)
if response.ok:
response_text = response.text
else:
return response.raise_for_status()
soup = BeautifulSoup(response_text, 'html.parser')
parent = [url + node.get('href') for node in soup.find_all('a') if node.get('href').endswith(ext)]
return parent
url = 'http://cdimage.debian.org/debian-cd/8.2.0-live/i386/iso-hybrid'
ext = 'iso'
result = get_url_paths(url, ext)
print(result)

HTTP does not work with "files" and "directories". Pick a different protocol.

You can use the following script to get names of all files in sub-directories and directories in a HTTP Server. A file writer can be used to download them.
from urllib.request import Request, urlopen, urlretrieve
from bs4 import BeautifulSoup
def read_url(url):
url = url.replace(" ","%20")
req = Request(url)
a = urlopen(req).read()
soup = BeautifulSoup(a, 'html.parser')
x = (soup.find_all('a'))
for i in x:
file_name = i.extract().get_text()
url_new = url + file_name
url_new = url_new.replace(" ","%20")
if(file_name[-1]=='/' and file_name[0]!='.'):
read_url(url_new)
print(url_new)
read_url("www.example.com")