I am trying to import a list of urls and grab pn2 and main1. I can run it without importing the file so I know it works but I just have no idea what to do with the import. Here is what I have tried most recent and below it is a small portion of the urls. Thanks in advance.
import urllib
import urllib.request
import csv
from bs4 import BeautifulSoup
csvfile = open("ecco1.csv")
csvfilelist = csvfile.read()
theurl="csvfilelist"
soup = BeautifulSoup(theurl,"html.parser")
for row in csvfilelist:
for pn in soup.findAll('td',{"class":"productText"}):
pn2.append(pn.text)
for main in soup.find_all('div',{"class":"breadcrumb"}):
main1 = main.text
print (main1)
print ('\n'.join(pn2))
Urls:
http://www.eccolink.com/products/productresults.aspx?catId=2458
http://www.eccolink.com/products/productresults.aspx?catId=2464
http://www.eccolink.com/products/productresults.aspx?catId=2435
http://www.eccolink.com/products/productresults.aspx?catId=2446
http://www.eccolink.com/products/productresults.aspx?catId=2463
From what I see, you are opening a CSV file and using BeautifulSoup to parse it.
That should not be the way.
BeautifulSoup parses html files, not CSV.
Looking at your code, it seems correct if you were passing in html code to Bs4.
from bs4 import BeautifulSoup
import requests
links = []
file = open('links.txt')
html = requests.get('http://www.example.com')
soup = BeautifulSoup(html, 'html.parser')
for x in soup.find_all('a',"class":"abc"):
links.append(x)
file.write(x)
file.close()
Above is a very basic implementation of how I could get a target element in the html code and write it to a file/ or append it to a list. Use Requests rather than urllib. It is a better library and more modern.
If you want to input your data as CSV, my best option is to use csv reader as import.
Hope that helps.
Related
I have written following code to open a local HTML file saved on my Desktop:
However while running this code I get following error:
I have no prior experience of handling this in Python or BS4. I tried various solutions online but couldn't solve it.
Code:
import csv
from email import header
from fileinput import filename
from tokenize import Name
import requests
from bs4 import BeautifulSoup
url = "C:\ Users\ ASUS\ Desktop\ payment.html"
page=open(url)
# r=requests.get(url)
# htmlContent = r.content
soup = BeautifulSoup(page.read())
head_tag = soup.head
for child in head_tag.descendants:
print(child)
Need help!
Thank you in advance.
It's unicode error prefix the path with r (to produce a raw string):
url = r"C:\ Users\ ASUS\ Desktop\ payment.html"
I'm trying to web scrape text and organize it, the text looks something like this:
https://i.stack.imgur.com/bKuXl.png
It is a mess.
How can I organize it to a json file or something?
Also I just did a basic web scrape:
from bs4 import BeautifulSoup
import requests
source = requests.get('https://api.hypixel.net/skyblock/bazaar').text
soup = BeautifulSoup(source, 'lxml')
print(soup.prettify())
Try this:
source = requests.get('https://api.hypixel.net/skyblock/bazaar')
json_response = source.json()
You can use the json module to dump into json format:
import json
import requests
source = requests.get('https://api.hypixel.net/skyblock/bazaar').text
json_file = json.dumps(source) # Returns a str() to json_file
# alternatively:
with open("path/to/file.txt", 'w') as file:
json.dump(source, file) # outputs json to file.txt
I am trying to scrape a page website/post-sitemap.xml which contains all url's posted for a wordpress website. In the first step, I need to make a list of all the url's present in post-sitemap. When I use requests.get and I check the output, it opens all of the internal urls as well, which is weird. My intention is to make a list of all url's first and then using a loop, I will scrape individual url's in the next function. Below is the code I have done so far. I would need all url's as a list as my final output if python gurus can help.
I have tried using requests.get and openurl but nothing seems to open only the base url for /post-sitemap.xml
import pandas as pd
import numpy as np
from urllib.request import urlopen
from bs4 import BeautifulSoup
import requests
import re
class wordpress_ext_url_cleanup(object):
def __init__(self,wp_url):
self.wp_url_raw = wp_url
self.wp_url = wp_url + '/post-sitemap.xml/'
def identify_ext_url(self):
html = requests.get(self.wp_url)
print(self.wp_url)
print(html.text)
soup = BeautifulSoup(html.text,'lxml')
#print(soup.get_text())
raw_data = soup.find_all('tr')
print (raw_data)
#for link in raw_data:
#print(link.get("href"))
def main():
print ("Inside Main Function");
url="http://punefirst dot com" #(knowingly removed the . so it doesnt look spammy)
first_call = wordpress_ext_url_cleanup(url)
first_call.identify_ext_url()
if __name__ == '__main__':
main()
I would need all 548 url's present in the post sitemap as a list which I will use it for the next function for further scraping.
The document that is returned from the server is XML and transformed with XSLT to HTML form (more info here). To parse all links from this XML, you can use this script:
import requests
from bs4 import BeautifulSoup
url = 'http://punefirst.com/post-sitemap.xml/'
soup = BeautifulSoup(requests.get(url).text, 'lxml')
for loc in soup.select('url > loc'):
print(loc.text)
Prints:
http://punefirst.com
http://punefirst.com/hospitals/pcmc-hospitals/aditya-birla-memorial-hospital-chinchwad-pune
http://punefirst.com/hospitals/pcmc-hospitals/saijyoti-hospital-and-icu-chinchwad-pune
http://punefirst.com/hospitals/pcmc-hospitals/niramaya-hospital-chinchwad-pune
http://punefirst.com/hospitals/pcmc-hospitals/chetna-hospital-chinchwad-pune
http://punefirst.com/hospitals/hadapsar-hospitals/pbmas-h-v-desai-eye-hospital
http://punefirst.com/hospitals/punecentral-hospitals/shree-sai-prasad-hospital
http://punefirst.com/hospitals/punecentral-hospitals/sadhu-vaswani-missions-medical-complex
http://punefirst.com/hospitals/katraj-kondhwa-hospitals/shivneri-hospital
http://punefirst.com/hospitals/punecentral-hospitals/kelkar-nursing-home
http://punefirst.com/hospitals/pcmc-hospitals/shrinam-hospital
http://punefirst.com/hospitals/pcmc-hospitals/dhanwantari-hospital-nigdi
http://punefirst.com/hospitals/punecentral-hospitals/dr-tarabai-limaye-hospital
http://punefirst.com/hospitals/katraj-kondhwa-hospitals/satyanand-hospital-kondhwa-pune
...and so on.
When I attempt to parse a locally stored copy of a webpage, beautifulsoup returns gibberish to me. I don't understand why as I've never faced this problem when using the requests and bs4 modules together for scraping tasks.
here's my code
import requests
from bs4 import BeautifulSoup as BS
import os
url_2 = r'/Users/davidferreira/Documents/coding_2/ak_screen_scraping/bmra/'
os.chdir(url_2)
f = open('re_2.html')
soup = BS(url_2, "lxml")
f.close()
print soup
this code returns the following :
<html><body><p>/Users/davidferreira/Documents/coding_2/ak_screen_scraping/bmra/</p></body></html>
I wasn't able to find a similar problem online so I've posted it here. any help would be much appreciated.
You are passing the path (which you named url_2) to BeautifulSoup so it treats that as a web page text and returns it, neatly wrapped in some minimal HTML. Seems fine.
Try constructing the BS from the file's contents instead. See here how it works: https://www.crummy.com/software/BeautifulSoup/bs4/doc/#making-the-soup
soup = BS(f)
should do...
I am trying to get all the urls on a website using python. At the moment I am just copying the websites html into the python program and then using code to extract all the urls. Is there a way I could do this straight from the web without having to copy the entire html?
In Python 2, you can use urllib2.urlopen:
import urllib2
response = urllib2.urlopen('http://python.org/')
html = response.read()
In Python 3, you can use urllib.request.urlopen:
import urllib.request
with urllib.request.urlopen('http://python.org/') as response:
html = response.read()
If you have to perform more complicated tasks like authentication or passing parameters I suggest to have a look at the requests library.
The most straightforward would probably be urllib.urlopen if you're using python2, or urllib.request.urlopen if you're using python3 (you have to do import urllib or import urllib.request first of course). That way you get an file like object from which you can read (ie f.read()) the html document.
Example for python 2:
import urllib
f = urlopen("http://stackoverflow.com")
http_document = f.read()
f.close()
The good news is that you seem to have done the hard part which is analyzing the html document for links.
You might want to use the bs4(BeautifulSoup) library.
Beautiful Soup is a Python library for pulling data out of HTML and XML files.
You can download bs4 with the followig command at the cmd line. pip install BeautifulSoup4
import urllib2
import urlparse
from bs4 import BeautifulSoup
url = "http://www.google.com"
response = urllib2.urlopen(url)
content = response.read()
soup = BeautifulSoup(content, "html.parser")
for link in soup.find_all('a', href=True):
print urlparse.urljoin(url, link['href'])
You can simply use the combination of requests and BeautifulSoup.
First make an HTTP request using requests to get the HTML content. You will get it as a Python string, which you can manipulate as you like.
Take the HTML content string and supply it into the BeautifulSoup, which has done all the job to extract the DOM, and get all URLs, i.e. <a> elements.
Here is an example of how to fetch all links from StackOverflow:
import requests
from bs4 import BeautifulSoup, SoupStrainer
response = requests.get('http://stackoverflow.com')
html_str = response.text
bs = BeautifulSoup(html_str, parseOnlyThese=SoupStrainer('a'))
for a_element in bs:
if a_element.has_attr('href'):
print(a_element['href'])
Sample output:
/questions/tagged/facebook-javascript-sdk
/questions/31743507/facebook-app-request-dialog-keep-loading-on-mobile-after-fb-login-called
/users/3545752/user3545752
/questions/31743506/get-nuspec-file-for-existing-nuget-package
/questions/tagged/nuget
...