So I have just started learning about python using the Coursera online course "Python for Everybody", and I have this assignment where I have to follow links using beautiful soup. I saw this question pop up before but when I tried using it, it just didn't work. I managed to create something but the thing doesn't actually follow through the links but instead just stays on the same page. If possible can anyone provide materials that can give better insight on this assignment as well? Thanks.
import urllib.request, urllib.parse, urllib.error
from bs4 import BeautifulSoup
import ssl
ctx = ssl.create_default_context()
ctx.check_hostname = False
ctx.verify_mode = ssl.CERT_NONE
url = input('Enter URL - ')
cnt = input("Enter count -")
count = int(cnt)
pn = input("Enter position -")
position = int(pn)-1
while count > 0:
html = urllib.request.urlopen(url, context=ctx).read()
soup = BeautifulSoup(html, "html.parser")
tags = soup('a')
lst = list()
for tag in tags:
lst.append(tag.get('href', None))
indxpos = lst[position]
count = count - 1
print("Retrieving:", indxpos)
You never set url to the new URL.
while count > 0:
html = urllib.request.urlopen(url, context=ctx).read() # Gets the page at url
...
for tag in tags:
lst.append(tag.get('href', None)) # Appends all the links to lst
indxpos = lst[position]
count = count - 1
print("Retrieving:", indxpos)
# What happens to lst?? you never use it
You should probably replace indxpos with url instead.
while count > 0:
html = urllib.request.urlopen(url, context=ctx).read() # Gets the page at url
...
for tag in tags:
lst.append(tag.get('href', None)) # Appends all the links to lst
url = lst[position]
count = count - 1
print("Retrieving:", url)
This way, the next time the loop runs, it will fetch the new URL.
Also: If the page does not have pn links (e.g. pn=12, page has 2 links), you will get an exception if you try and access lst[position], because lst has less than pn elements.
You don't have a function that interacts with the list of hyperlinks in your code, whatsoever.
It will only print contents of "lst" list, but won't do anything with them.
A am learning how to use BeautifulSoup and I have run into an issue with double printing in a loop I have written.
Any insight would be greatly appreciated!
from bs4 import BeautifulSoup
import requests
import re
page = 'https://news.google.com/news/headlines?gl=US&ned=us&hl=en' #main page
#url = raw_input("Enter a website to extract the URL's from: ")
r = requests.get(page) #requests html document
data = r.text #set data = to html text
soup = BeautifulSoup(data, "html.parser") #parse data with BS
for link in soup.find_all('a'):
#if contains /news/
if ('/news/' in link.get('href')):
print(link.get('href'))
Examples:
for link in soup.find_all('a'):
#if contains cointelegraph/news/
#if ('https://cointelegraph.com/news/' in link.get('href')):
url = link.get('href') #local var store url
if '/news/' in url:
print(url)
print(count)
count += 1
if count == 5:
break
output:
https://cointelegraph.com/news/woman-in-denmark-imprisoned-for-hiring-hitman-using-bitcoin
0
https://cointelegraph.com/news/ethereum-price-hits-all-time-high-of-750-following-speed-boost
1
https://cointelegraph.com/news/ethereum-price-hits-all-time-high-of-750-following-speed-boost
2
https://cointelegraph.com/news/senior-vp-says-ebay-seriously-considering-bitcoin-integration
3
https://cointelegraph.com/news/senior-vp-says-ebay-seriously-considering-bitcoin-integration
4
For some reason my code keeps printing out the same url twice...
Based on your code and the provided link there seems to be duplicates in the results of BeautifulSoup find_all search. The html structure needs to be checked to see why duplicates are returned (check the find_all search options to filter some in the documentation. But if you want a quick fix and want to remove the duplicates from the printed results you can use the modified loop with a set as below to keep track of seen entries (based on this).
In [78]: l = [link.get('href') for link in soup.find_all('a') if '/news/' in link.get('href')]
In [79]: any(l.count(x) > 1 for x in l)
Out[79]: True
The above output shows duplicate exists in the list. Now to remove them use something like
seen = set()
for link in soup.find_all('a'):
lhref = link.get('href')
if '/news/' in lhref and lhref not in seen:
print lhref
seen.add(lhref)
I am trying to create a webcrawler that parses all the html on the page, grabs a specified (via raw_input) link, follows that link, and then repeats this process a specified number of times (once again via raw_input). I am able to grab the first link and successfully print it. However, I am having problems "looping" the whole process, and usually grab the wrong link. This is the first link
https://pr4e.dr-chuck.com/tsugi/mod/python-data/data/known_by_Fikret.html
(Full disclosure, this questions pertains to an assignment for a Coursera course)
Here's my code
import urllib
from BeautifulSoup import *
url = raw_input('Enter - ')
rpt=raw_input('Enter Position')
rpt=int(rpt)
cnt=raw_input('Enter Count')
cnt=int(cnt)
count=0
counts=0
tags=list()
soup=None
while x==0:
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html)
# Retrieve all of the anchor tags
tags=soup.findAll('a')
for tag in tags:
url= tag.get('href')
count=count + 1
if count== rpt:
break
counts=counts + 1
if counts==cnt:
x==1
else: continue
print url
Based on DJanssens' response, I found the solution;
url = tags[position-1].get('href')
did the trick for me!
Thanks for the assistance!
I also worked on that course, and help with a friend, I got this worked out:
import urllib
from bs4 import BeautifulSoup
url = "http://python-data.dr-chuck.net/known_by_Happy.html"
rpt=7
position=18
count=0
counts=0
tags=list()
soup=None
x=0
while x==0:
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html,"html.parser")
tags=soup.findAll('a')
url= tags[position-1].get('href')
count=count + 1
if count == rpt:
break
print url
I believe this is what you are looking for:
import urllib
from bs4 import *
url = raw_input('Enter - ')
position=int(raw_input('Enter Position'))
count=int(raw_input('Enter Count'))
#perform the loop "count" times.
for _ in xrange(0,count):
html = urllib.urlopen(url).read()
soup = BeautifulSoup(html)
tags=soup.findAll('a')
for tag in tags:
url= tag.get('href')
tags=soup.findAll('a')
# if the link does not exist at that position, show error.
if not tags[position-1]:
print "A link does not exist at that position."
# if the link at that position exist, overwrite it so the next search will use it.
url = tags[position-1].get('href')
print url
The code will now loop the amount of times as specified in the input, each time it will take the href at the given position and replace it with the url, in that way each loop will look further in the tree structure.
I advice you to use full names for variables, which is a lot easier to understand. In addition you could cast them and read them in a single line, which makes your beginning easier to follow.
Here is my 2-cents:
import urllib
#import ssl
from bs4 import BeautifulSoup
#'http://py4e-data.dr-chuck.net/known_by_Fikret.html'
url = raw_input('Enter URL : ')
position = int(raw_input('Enter position : '))
count = int(raw_input('Enter count : '))
print('Retrieving: ' + url)
soup = BeautifulSoup(urllib.urlopen(url).read())
for x in range(1, count + 1):
link = list()
for tag in soup('a'):
link.append(tag.get('href', None))
print('Retrieving: ' + link[position - 1])
soup = BeautifulSoup(urllib.urlopen(link[position - 1]).read())
I am trying to scrape data from the PGA.com website to get a table of all of the golf courses in the United States. In my CSV table I want to include the Name of the golf course ,Address ,Ownership ,Website , Phone number. With this data I would like to geocode it and place into a map and have a local copy on my computer
I utilized Python and Beautiful Soup4 to extract my data. I have reached as far to extract the data and import it into a CSV but I am now having a problem of scraping data from multiple pages on the PGA website. I want to extract ALL THE GOLF COURSES but my script is limited only to one page I want to loop it in away that it will capture all data for golf courses from all pages found in the PGA site. There are about 18000 gold courses and 900 pages to capture data
Attached below is my script. I need help on creating code that will capture ALL data from the PGA website and not just one site but multiple. In this manner it will provide me with all the data of gold courses in the United States.
Here is my script below:
import csv
import requests
from bs4 import BeautifulSoup
url = "http://www.pga.com/golf-courses/search?searchbox=Course+Name&searchbox_zip=ZIP&distance=50&price_range=0&course_type=both&has_events=0"
r = requests.get(url)
soup = BeautifulSoup(r.content)
g_data1=soup.find_all("div",{"class":"views-field-nothing-1"})
g_data2=soup.find_all("div",{"class":"views-field-nothing"})
courses_list=[]
for item in g_data2:
try:
name=item.contents[1].find_all("div",{"class":"views-field-title"})[0].text
except:
name=''
try:
address1=item.contents[1].find_all("div",{"class":"views-field-address"})[0].text
except:
address1=''
try:
address2=item.contents[1].find_all("div",{"class":"views-field-city-state-zip"})[0].text
except:
address2=''
try:
website=item.contents[1].find_all("div",{"class":"views-field-website"})[0].text
except:
website=''
try:
Phonenumber=item.contents[1].find_all("div",{"class":"views-field-work-phone"})[0].text
except:
Phonenumber=''
course=[name,address1,address2,website,Phonenumber]
courses_list.append(course)
with open ('filename5.csv','wb') as file:
writer=csv.writer(file)
for row in courses_list:
writer.writerow(row)
#for item in g_data1:
#try:
#print item.contents[1].find_all("div",{"class":"views-field-counter"})[0].text
#except:
#pass
#try:
#print item.contents[1].find_all("div",{"class":"views-field-course-type"})[0].text
#except:
#pass
#for item in g_data2:
#try:
#print item.contents[1].find_all("div",{"class":"views-field-title"})[0].text
#except:
#pass
#try:
#print item.contents[1].find_all("div",{"class":"views-field-address"})[0].text
#except:
#pass
#try:
#print item.contents[1].find_all("div",{"class":"views-field-city-state-zip"})[0].text
#except:
#pass
This script only captures 20 at a time and I want to capture all in one script which account for 18000 golf courses and 900 pages to scrape form.
The PGA website's search have multiple pages, the url follows the pattern:
http://www.pga.com/golf-courses/search?page=1 # Additional info after page parameter here
this means you can read the content of the page, then change the value of page by 1, and read the the next page.... and so on.
import csv
import requests
from bs4 import BeautifulSoup
for i in range(907): # Number of pages plus one
url = "http://www.pga.com/golf-courses/search?page={}&searchbox=Course+Name&searchbox_zip=ZIP&distance=50&price_range=0&course_type=both&has_events=0".format(i)
r = requests.get(url)
soup = BeautifulSoup(r.content)
# Your code for each individual page here
if you still read this post , you can try this code too....
from urllib.request import urlopen
from bs4 import BeautifulSoup
file = "Details.csv"
f = open(file, "w")
Headers = "Name,Address,City,Phone,Website\n"
f.write(Headers)
for page in range(1,5):
url = "http://www.pga.com/golf-courses/search?page={}&searchbox=Course%20Name&searchbox_zip=ZIP&distance=50&price_range=0&course_type=both&has_events=0".format(page)
html = urlopen(url)
soup = BeautifulSoup(html,"html.parser")
Title = soup.find_all("div", {"class":"views-field-nothing"})
for i in Title:
try:
name = i.find("div", {"class":"views-field-title"}).get_text()
address = i.find("div", {"class":"views-field-address"}).get_text()
city = i.find("div", {"class":"views-field-city-state-zip"}).get_text()
phone = i.find("div", {"class":"views-field-work-phone"}).get_text()
website = i.find("div", {"class":"views-field-website"}).get_text()
print(name, address, city, phone, website)
f.write("{}".format(name).replace(",","|")+ ",{}".format(address)+ ",{}".format(city).replace(",", " ")+ ",{}".format(phone) + ",{}".format(website) + "\n")
except: AttributeError
f.close()
where it is written range(1,5) just change that with 0,to the last page , and you will get all details in CSV, i tried very hard to get your data in proper format but it's hard:).
You're putting a link to a single page, it's not going to iterate through each one on its own.
Page 1:
url = "http://www.pga.com/golf-courses/search?searchbox=Course+Name&searchbox_zip=ZIP&distance=50&price_range=0&course_type=both&has_events=0"
Page 2:
http://www.pga.com/golf-courses/search?page=1&searchbox=Course%20Name&searchbox_zip=ZIP&distance=50&price_range=0&course_type=both&has_events=0
Page 907:
http://www.pga.com/golf-courses/search?page=906&searchbox=Course%20Name&searchbox_zip=ZIP&distance=50&price_range=0&course_type=both&has_events=0
Since you're running for page 1 you'll only get 20. You'll need to create a loop that'll run through each page.
You can start off by creating a function that does one page then iterate that function.
Right after the search? in the url, starting at page 2, page=1 begins increasing until page 907 where it's page=906.
I noticed that the first solution had a repetition of the first instance, that is because the 0 page and 1 page is the same page. This is resolved by specifying the start page in the range function. Example below...
for i in range(1, 907): #Number of pages plus one
url = "http://www.pga.com/golf-courses/search?page={}&searchbox=Course+Name&searchbox_zip=ZIP&distance=50&price_range=0&course_type=both&has_events=0".format(i)
r = requests.get(url)
soup = BeautifulSoup(r.content, "html5lib") #Can use whichever parser you prefer
# Your code for each individual page here
Had this same exact problem and the solutions above did not work. I solved mine by accounting for cookies. A requests session helps. Create a session and it'll pull all the pages you need by inserting a cookie to all the numbered pages.
import csv
import requests
from bs4 import BeautifulSoup
url = "http://www.pga.com/golf-courses/search?searchbox=Course+Name&searchbox_zip=ZIP&distance=50&price_range=0&course_type=both&has_events=0"
s = requests.Session()
r = s.get(url)
The PGA website has changed this question has been asked.
It seems they organize all courses by: State > City > Course
In light of this change and the popularity of this question, here's how I'd solve this problem today.
Step 1 - Import everything we'll need:
import time
import random
from gazpacho import Soup # https://github.com/maxhumber/gazpacho
from tqdm import tqdm # to keep track of progress
Step 2 - Scrape all the state URL endpoints:
URL = "https://www.pga.com"
def get_state_urls():
soup = Soup.get(URL + "/play")
a_tags = soup.find("ul", {"data-cy": "states"}, mode="first").find("a")
state_urls = [URL + a.attrs['href'] for a in a_tags]
return state_urls
state_urls = get_state_urls()
Step 3 - Write a function to scrape all the city links:
def get_state_cities(state_url):
soup = Soup.get(state_url)
a_tags = soup.find("ul", {"data-cy": "city-list"}).find("a")
state_cities = [URL + a.attrs['href'] for a in a_tags]
return state_cities
state_url = state_urls[0]
city_links = get_state_cities(state_url)
Step 4 - Write a function to scrape all of the courses:
def get_courses(city_link):
soup = Soup.get(city_link)
courses = soup.find("div", {"class": "MuiGrid-root MuiGrid-item MuiGrid-grid-xs-12 MuiGrid-grid-md-6"}, mode="all")
return courses
city_link = city_links[0]
courses = get_courses(city_link)
Step 5 - Write a function to parse all the useful info about a course:
def parse_course(course):
return {
"name": course.find("h5", mode="first").text,
"address": course.find("div", {'class': "jss332"}, mode="first").strip(),
"url": course.find("a", mode="first").attrs["href"]
}
course = courses[0]
parse_course(course)
Step 6 - Loop through everything and save:
all_courses = []
for state_url in tqdm(state_urls):
city_links = get_state_cities(state_url)
time.sleep(random.uniform(1, 10) / 10)
for city_link in city_links:
courses = get_courses(city_link)
time.sleep(random.uniform(1, 10) / 10)
for course in courses:
info = parse_course(course)
all_courses.append(info)
I'm trying to scrape a list of URL's from the European Parliament's Legislative Observatory. I do not type in any search keyword in order to get all links to documents (currently 13172). I can easily scrape a list of the first 10 results which are displayed on the website using the code below. However, I want to have all links so that I would not need to somehow press the next page button. Please let me know if you know of a way to achieve this.
import requests, bs4, re
# main url of the Legislative Observatory's search site
url_main = 'http://www.europarl.europa.eu/oeil/search/search.do?searchTab=y'
# function gets a list of links to the procedures
def links_to_procedures (url_main):
# requesting html code from the main search site of the Legislative Observatory
response = requests.get(url_main)
soup = bs4.BeautifulSoup(response.text) # loading text into Beautiful Soup
links = [a.attrs.get('href') for a in soup.select('div.procedure_title a')] # getting a list of links of the procedure title
return links
print(links_to_procedures(url_main))
You can follow the pagination by specifying the page GET parameter.
First, get the results count, then calculate the number of pages to process by dividing the count on the results count per page. Then, iterate over pages one by one and collect the links:
import re
from bs4 import BeautifulSoup
import requests
response = requests.get('http://www.europarl.europa.eu/oeil/search/search.do?searchTab=y')
soup = BeautifulSoup(response.content)
# get the results count
num_results = soup.find('span', class_=re.compile('resultNum')).text
num_results = int(re.search('(\d+)', num_results).group(1))
print "Results found: " + str(num_results)
results_per_page = 50
base_url = "http://www.europarl.europa.eu/oeil/search/result.do?page={page}&rows=%s&sort=d&searchTab=y&sortTab=y&x=1411566719001" % results_per_page
links = []
for page in xrange(1, num_results/results_per_page + 1):
print "Current page: " + str(page)
url = base_url.format(page=page)
response = requests.get(url)
soup = BeautifulSoup(response.content)
links += [a.attrs.get('href') for a in soup.select('div.procedure_title a')]
print links