Downloading files from web [closed] - python

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i need first 5 text files from URL using python: http://www.textfiles.com/etext/AUTHORS/SHAKESPEARE/ , only '.txt' files should be dowloaded and store it in folder
I tried using requests lib to get access to website

Here I've done it using requests to get URLs content, and BeautifulSoup to retrieve urls to download .txt's from main page
Download page content using requests
Using BeautifulSoup, find all <a> tags
Get first 5 tags that ends up with .txt
Download content of those tags href using requests
import requests
from bs4 import BeautifulSoup
url = "http://www.textfiles.com/etext/AUTHORS/SHAKESPEARE/"
AMOUNT_OF_FILES = 5 # Amount of txt files to download
FILES_EXTENSION = ".txt" # Extension to download
# Getting url content
req = requests.get(url)
soup = BeautifulSoup(req.text, "html.parser")
# Finding all a tags in table
a_tags = soup.find("table").find_all("a")
urls_to_download = []
# Getting urls to download .txt`s
for a_tag in a_tags:
if a_tag['href'].endswith(FILES_EXTENSION):
urls_to_download.append(url + a_tag['href'])
if len(urls_to_download) == AMOUNT_OF_FILES:
break
# Downloading file contents
for url in urls_to_download:
filename = url[url.rindex("/")+1:]
request_url = requests.get(url)
with open(filename, "wb") as file:
file.write(request_url.content)

Related

Getting Text value from element using beautifulsoup in Python [closed]

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I am making python script which getting text data from online site.
this is simple web scraping script and the language is only python.
I don't use selenium and use only beautifulsoup.
and I can scrape text from <p> or <div> or even <h> and <a>
but when I try to get text from <td>, the code is not working.
I shared my code below.
from threading import Thread
from bs4 import BeautifulSoup
from lxml import etree
detailPage = requests.get(SUBURL, headers=HEADERS)
detailsoup = BeautifulSoup(detailPage.content, "html.parser")
detaildom = etree.HTML(str(detailsoup))
name = detaildom.xpath('//*[#id="productTitle"]')[0].text
asin = detaildom.xpath('//*[#id="productDetails_detailBullets_sections1"]/tbody/tr[1]/td')[0].text
here, getting name is working, asin return empty string.
You can find the table by its ID productDetails_detailBullets_sections1 and find the <td> which contains the "ASIN".
Using a CSS selector:
soup = BeautifulSoup(requests.get(url, headers=headers).content, "html.parser")
print("ASIN:", soup.select_one("#productDetails_db_sections tr > td").get_text(strip=True))
Using .find():
soup = BeautifulSoup(requests.get(url, headers=headers).content, "html.parser")
table_info = soup.find(id="productDetails_detailBullets_sections1").find("tr")
print("ASIN:", table_info.find('td').get_text(strip=True))
Output (in both solutions):
ASIN: B079LWYC17

HTTP Error 403: Forbidden with Tabula/Requests [closed]

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I am getting the error "urllib.error.HTTPError: HTTP Error 403: Forbidden" with Tabula, is there a way to fix this? It has worked correctly for most of this year:
import tabula
from bs4 import BeautifulSoup
import requests
url = 'https://www.who.int/emergencies/diseases/novel-coronavirus-2019/situation-reports'
r = requests.get(url)
soup = BeautifulSoup(r.content, 'lxml' )
hyperlink_tags = soup.find_all('a')
for hyperlink_tag in hyperlink_tags:
if 'Situation report' in hyperlink_tag.text:
file_path = hyperlink_tag['href']
break
latest_report = f'https://who.int/{file_path}'
file = latest_report
tables = tabula.read_pdf(file, stream=True, pages = "all", multiple_tables = True)
The problem seems to be the last line so I'm not sure if it's requests or tabula
the request needs the headers parameter for the User-Agent. not sure how to add that parameter with tabula, but you can access and write the pdf to file, then read that in:
import tabula
from bs4 import BeautifulSoup
import requests
url = 'https://www.who.int/emergencies/diseases/novel-coronavirus-2019/situation-reports'
r = requests.get(url)
soup = BeautifulSoup(r.content, 'lxml' )
hyperlink_tags = soup.find_all('a')
for hyperlink_tag in hyperlink_tags:
if 'Situation report' in hyperlink_tag.text:
file_path = hyperlink_tag['href']
break
latest_report = f'https://who.int/{file_path}'
file = latest_report
################################################
## Download the PDF ############################
from urllib.request import Request, urlopen
f = open('c:/test/temp.pdf', 'wb')
url_request = Request(file,
headers={"User-Agent": "Mozilla/5.0"})
webpage = urlopen(url_request).read()
f.write(webpage)
f.close()
#################################################
tables = tabula.read_pdf('c:/test/temp.pdf', stream=False, pages = "all", multiple_tables = True)

in python, I have 100 page's link and want to save them as html [closed]

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100 page's links inside (links.txt)
This is the code I have so far (it is save only one page) but the part of saving all the 99 pages is missing
import requests
import urllib.request, urllib.error, urllib.parse
with open('links.txt', 'r') as links:
for link in links:
response = urllib.request.urlopen(link)
webContent = response.read()
f = open('obo-t17800628-33.html', 'wb')
f.write(webContent)
f.close
You need to give the files different names as you loop:
import requests
import urllib.request, urllib.error, urllib.parse
with open('links.txt', 'r') as links:
for idx, link in enumerate(links):
response = urllib.request.urlopen(link)
webContent = response.read()
with open('obo-t17800628-33.html' + str(idx), 'wb') as fout:
fout.write(webContent)
This will append a number to the end of each file name.

How extract URL from web page [closed]

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I need a way to extract url from the list at this web page https://iota-nodes.net/
using Python. I tried BeautifulSoup but without success.
My code is:
from bs4 import BeautifulSoup, SoupStrainer
import requests
url = "https://iota-nodes.net/"
page = requests.get(url)
data = page.text
soup = BeautifulSoup(data)
for link in soup.find_all('a'):
print(link.get('href'))
No need for BeautifulSoup, as the data is coming from an AJAX request. Something like this should work:
import requests
response = requests.get('https://api.iota-nodes.net/')
data = response.json()
hostnames = [node['hostname'] for node in data]
Note that the data comes from an API endpoint being https://api.iota-nodes.net/.

Find hyperlinks of a page in python without BeautifulSoup [closed]

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What I am trying to do is find all the hyper links of a web page here is what I have so far but it does not work
from urllib.request import urlopen
def findHyperLinks(webpage):
link = "Not found"
encoding = "utf-8"
for webpagesline in webpage:
webpagesline = str(webpagesline, encoding)
if "<a href>" in webpagesline:
indexstart = webpagesline.find("<a href>")
indexend = webpagesline.find("</a>")
link = webpagesline[indexstart+7:indexend]
return link
return link
def main():
address = input("Please enter the adress of webpage to find the hyperlinks")
try:
webpage = urlopen(address)
link = findHyperLinks(webpage)
print("The hyperlinks are", link)
webpage.close()
except Exception as exceptObj:
print("Error:" , str(exceptObj))
main()
There are multiple problems in your code. One of them is that you are trying to find links with present, empty and the only one href attribute: <a href>.
Anyway, if you would use an HTML parser (well, to parse HTML), things would get much more easy and reliable. Example using BeautifulSoup:
from bs4 import BeautifulSoup
from urllib.request import urlopen
soup = BeautifulSoup(urlopen(address))
for link in soup.find_all("a", href=True):
print(link["href"], link.get_text())
Without BeautifulSoap you can use RegExp and simple function.
from urllib.request import urlopen
import re
def find_link(url):
response = urlopen(url)
res = str(response.read())
my_dict = re.findall('(?<=<a href=")[^"]*', res)
for x in my_dict:
# simple skip page bookmarks, like #about
if x[0] == '#':
continue
# simple control absolute url, like /about.html
# also be careful with redirects and add more flexible
# processing, if needed
if x[0] == '/':
x = url + x
print(x)
find_link('http://cnn.com')

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