Develop Reference

How to find the total number of pages on a website with BeautifulSoup?

Context: I'm working on pagination of this website: https://skoodos.com/schools-in-uttarakhand. When I inspected, this website has no proper number of pages visible except the next button which is ?page=2 after the url. Also, searching for page-link gave me number 20 at the end. So I assumed that the total number of pages is 20, upon checking manually, I learnt that, there only exist 11 pages.
After many trials and errors, I finally decided to go with just the indexing from 0 until 12 (12 is excluded by python however).
What I want to know is that, how wold you go about figuring out the number of pages on a particular website that doesn't show the actual number of pages other than previous and next button and how can I optimize this in terms of the same?
Here's my solution to pagination. Any way to optimize this other than me manually finding the number of pages?
from myWork.commons import url_parser, write
def data_fetch(url):
school_info = []
for page_number in range(0, 4):
next_web_page = url + f'?page={page_number}'
soup = url_parser(next_web_page)
search_results = soup.find('section', {'id': 'search-results'}).find(class_='container').find(class_='row')
# rest of the code
for page_number in range(4, 12):
next_web_page = url + f'?page={page_number}'
soup = url_parser(next_web_page)
search_results = soup.find('section', {'id': 'search-results'}).find(class_='container').find(class_='row')
# rest of the code
def main():
url = "https://skoodos.com/schools-in-uttarakhand"
data_fetch(url)
if __name__ == "__main__":
main()

Each of your pages (except the last one) will have an element like this:
<a class="page-link"
href="https://skoodos.com/schools-in-uttarakhand?page=2"
rel="next">Next »</a>
E.g. you can extract the link as follows (here for the first page):
link = soup.find('a', class_='page-link', href=True, rel='next')
print(link['href'])
https://skoodos.com/schools-in-uttarakhand?page=2
So, you could make your function recursive. E.g. use something like this:
import requests
from bs4 import BeautifulSoup
def data_fetch(url, results = list()):
resp = requests.get(url)
soup = BeautifulSoup(resp.content, 'lxml')
search_results = soup.find('section', {'id': 'search-results'})\
.find(class_='container').find(class_='row')
results.append(search_results)
link = soup.find('a', class_='page-link', href=True, rel='next')
# link will be `None` for last page (i.e. `page=11`)
if link:
# just adding some prints to show progress of iteration
if not 'page' in url:
print('getting page: 1', end=', ')
url = link['href']
# subsequent page nums being retrieved
print(f'{url.rsplit("=", maxsplit=1)[1]}', end=', ')
# recursive call
return data_fetch(url, results)
else:
# `page=11` with no link, we're done
print('done')
return results
url = 'https://skoodos.com/schools-in-uttarakhand'
data = data_fetch(url)
So, a call to this function will print progress as:
getting page: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, done
And you'll end up with data with 11x bs4.element.Tag, one for each page.
print(len(data))
11
print(set([type(d) for d in data]))
{<class 'bs4.element.Tag'>}
Good luck with extracting the required info; the site is very slow, and the HTML is particularly sloppy and inconsistent. (e.g. you're right to note there is a page-link elem, which suggests there are 20 pages. But its visibility is set to hidden, so apparently this is just a piece of deprecated/unused code.)

There's a bit at the top that says "Showing the 217 results as per selected criteria". You can code to extract the number from that, then count the number number of results per page and divide by that to get the expected number of pages (don't forget to round up ).
If you want to double check, add more code to go to the calculated last page and
if there's no such page, keep decrementing the total and checking until you hit a page that exists
if there is such a page, but it has an active/enabled "Next" button, keep going to Next page until reaching the last (basically as you are now)
(Remember that the two listed above are contingencies and wouldn't be executed in an ideal scenario.)
So, just to find the number of pages, you could do something like:
import requests
from bs4 import BeautifulSoup
import math
def soupFromUrl(scrapeUrl):
req = requests.get(scrapeUrl)
if req.status_code == 200:
return BeautifulSoup(req.text, 'html.parser')
else:
raise Exception(f'{req.reason} - failed to scrape {scrapeUrl}')
def getPageTotal(url):
soup = soupFromUrl(url)
#totalResults = int(soup.find('label').get_text().split('(')[-1].split(')')[0])
totalResults = int(soup.p.strong.get_text()) # both searches should work
perPageResults = len(soup.select('.m-show')) #probably always 20
print(f'{perPageResults} of {totalResults} results per page')
if not (perPageResults > 0 and totalResults > 0):
return 0
lastPageNum = math.ceil(totalResults/perPageResults)
# Contingencies - will hopefully never be needed
lpSoup = soupFromUrl(f'{url}?page={lastPageNum}')
if lpSoup.select('.m-show'): #page exists
while lpSoup.select_one('a[rel="next"]'):
nextLink = lpSoup.select_one('a[rel="next"]')['href']
lastPageNum = int(nextLink.split('page=')[-1])
lpSoup = soupFromUrl(nextLink)
else: #page does not exist
while not (lpSoup.select('.m-show') or lastPageNum < 1):
lastPageNum = lastPageNum - 1
lpSoup = soupFromUrl(f'{url}?page={lastPageNum}')
# end Contingencies section
return lastPageNum
However, it looks like you only want the total pages in order to start the for-loop, but it's not even necessary to use a for-loop at all - a while-loop might be better:
def data_fetch(url):
school_info = []
nextUrl = url
while nextUrl:
soup = soupFromUrl(nextUrl)
#GET YOUR DATA FROM PAGE
nextHL = soup.select_one('a[rel="next"]')
nextUrl = nextHL.get('href') if nextHL else None
# code after fetching all pages' data
Although, you could still use for-loop if you had a maximum page number in mind:
def data_fetch(url, maxPages):
school_info = []
for p in range(1, maxPages+1):
soup = soupFromUrl(f'{url}?page={p}')
if not soup.select('.m-show'):
break
#GET YOUR DATA FROM PAGE
# code after fetching all pages' data [upto max]

Stuck Scraping with Beautifulsoup

So i'm trying to scrape a html webpage. It has novel chapters and i'm trying to get the text and store in text files to read offline. I don't have any previous experience with html or other things either. So the webpage I am trying to scrape is this. And the code i've been testing so far looks like this
`
import sys
import requests
import time
import re
from bs4 import BeautifulSoup
def browse_and_scrape(seed_url, page_number=1):
# Fetch the URL - We will be using this to append to images and info routes
url_pat = re.compile(r"(http://.*\.org)")
source_url = url_pat.search(seed_url).group(0)
# Page_number from the argument gets formatted in the URL & Fetched
formatted_url = seed_url.format(str(page_number))
# print(url_pat,source_url,formatted_url)
try:
html_text = requests.get(formatted_url).text
# print(html_text)
# Prepare the soup
soup = BeautifulSoup(html_text, "html.parser")
print(soup.find_all(id="chapterContent")[0]["style"])
print(f"Now Scraping - {formatted_url}")
# help = soup.find_all("div",class_="chapter-content text-normal")[0].text.strip().encode("ascii", "ignore").decode("ascii")
# for node in soup.findAll("div",class_="chapter-content text-normal"):
# print(node)
# print(''.join(node.findAll(text=True)))
# for node in soup.findAll("div"):
# # print(node)
# print(''.join(node.findAll(text=True)))
# help = soup.find_all("div",class_="chapter-content text-normal")[0]
# print(''.join(help.findAll(text=True)))
# print(help)
except Exception as e:
return e
return true
if __name__ == "__main__":
# seed_url = "http://books.toscrape.com/catalogue/page-{}.html"
seed_url = "http://wnmtl.org/chapter/324909-heavenly-wolf-valley.html"
# seed_url = "http://wnmtl.org/chapter/{}.html"
print("Web scraping has begun")
result = browse_and_scrape(seed_url)
if result == True:
print("Web scraping is now complete!")
else:
print(f"Oops, That doesn't seem right!!! - {result}")`
All the commented stuff are things i've been trying to rip the text from the tag. From my inspection of the developer console in the browser, all the text is in the tag with id of chapter content. My plan is to iteratively get the text, stuff it, get the link for the next page and repeat but i've been stuck for a bit now, any suggestions.

Instead of scraping each page, you can directly get the text from this API endpoint using requests.
https://api.mystorywave.com/story-wave-backend/api/v1/content/chapters/324909
The last item in the above API is the chapter ID (324909). You can navigate to chapters by giving in the chapter IDs.
The next and prev chapter IDs are present in the current chapter's API endpoint. Have a look at the above URL in browser to understand it better.
Here is the full recursive code that writes the text from 3 pages to a file called novel.txt. You may change the number of pages and other details as per your need.
import requests
def get_data(chapter_id, pages):
if pages == 0:
return
url = 'https://api.mystorywave.com/story-wave-backend/api/v1/content/chapters/' + str(chapter_id)
r = requests.get(url)
x = r.json()
pre_id = x['data']['preId']
next_id = x['data']['nextId']
title = x['data']['title']
content = x['data']['content']
chapter_title = f'\n***** Chapter: {title} *****\n'
with open('novel.txt', 'a') as f:
f.write(chapter_title)
f.write(content + '\n')
print(f"Chapter: '{title}' written to file.")
get_data(next_id, pages-1)
curr_id = '324909'
get_data(curr_id, 3)
Chapter: 'Heavenly Wolf Valley' written to file.
Chapter: 'Leaving' written to file.
Chapter: 'Pure Fabrication' written to file.

Using BeautifulSoup to scrape li's and id's in same method

How would i modify the parameters of the findAll method to read both li's and id's? li's are elements and id's are attributes correct?
#Author: David Owens
#File name: soupScraper.py
#Description: html scraper that takes surf reports from various websites
import csv
import requests
from bs4 import BeautifulSoup
###################### SURFLINE URL STRINGS AND TAG ###########################
slRootUrl = 'http://www.surfline.com/surf-report/'
slSunsetCliffs = 'sunset-cliffs-southern-california_4254/'
slScrippsUrl = 'scripps-southern-california_4246/'
slBlacksUrl = 'blacks-southern-california_4245/'
slCardiffUrl = 'cardiff-southern-california_4786/'
slTagText = 'observed-wave-range'
slTag = 'id'
#list of surfline URL endings
slUrls = [slSunsetCliffs, slScrippsUrl, slBlacksUrl, slCardiffUrl]
###############################################################################
#################### MAGICSEAWEED URL STRINGS AND TAG #########################
msRootUrl = 'http://magicseaweed.com/'
msSunsetCliffs = 'Sunset-Cliffs-Surf-Report/4211/'
msScrippsUrl = 'Scripps-Pier-La-Jolla-Surf-Report/296/'
msBlacksUrl = 'Torrey-Pines-Blacks-Beach-Surf-Report/295/'
msTagText = 'rating-text text-dark'
msTag = 'li'
#list of magicseaweed URL endings
msUrls = [msSunsetCliffs, msScrippsUrl, msBlacksUrl]
###############################################################################
'''
This method iterates through a list of urls and extracts the surf report from
the webpage dependent upon its tag location
rootUrl: The root url of each surf website
urlList: A list of specific urls to be appended to the root url for each
break
tag: the html tag where the actual report lives on the page
returns: a list of strings of each breaks surf report
'''
def extract_Reports(rootUrl, urlList, tag, tagText):
#empty list to hold reports
reports = []
#loop thru URLs
for url in urlList:
try:
#request page
request = requests.get(rootUrl + url)
#turn into soup
soup = BeautifulSoup(request.content, 'lxml')
#get the tag where report lives
reportTag = soup.findAll(id = tagText)
for report in reportTag:
reports.append(report.string.strip())
#notify if fail
except:
print 'scrape failure'
pass
return reports
#END METHOD
slReports = extract_Reports(slRootUrl, slUrls, slTag, slTagText)
msReports = extract_Reports(msRootUrl, msUrls, msTag, msTagText)
print slReports
print msReports
As of right now, only slReports prints correctly because i have it explicitly set to id = tagText. I am also aware that my tag paramater is not used currently.

So the problem is that you want to search the parse tree for elements that have either a class name of rating-text (it turns out you do not need text-dark to identify the relevant elements in the case of Magicseaweed) or an ID of observed-wave-range, using a single findAll call.
You can use a filter function to achieve this:
def reportTagFilter(tag):
return (tag.has_attr('class') and 'rating-text' in tag['class']) \
or (tag.has_attr('id') and tag['id'] == 'observed-wave-range')
Then change your extract_Reports function to read:
reportTag = soup.findAll(reportTagFilter)[0]
reports.append(reportTag.text.strip())

Python flask-paginate returning too many results on page

Im using the module flask-paginate found here: http://pythonhosted.org/Flask-paginate/
I am able to return results, and my pagination begins with the correct number of pages, but its currently showing 69 results per page. The documentation shows per_page= , but this only affects my starting page number which is correct. Im using SQL-Alchemy for my db.
#search.route('/search')
def search():
page, per_page, offset, inner_window = get_page_items()
links = Item.query.all()
total = Item.query.count()
pagination = get_pagination(page=page,
per_page=per_page,
total = total,
format_total=True,
format_number=True,
record_name='links',
)
return render_template('search/searchPage.html', offset=offset, total=total, links=links, pagination=pagination, per_page=per_page, page=page)
def get_css_framework():
return 'bootstrap3'
def get_link_size():
return 'sm' #option lg
def show_single_page_or_not():
return False
def get_page_items():
page = int(request.args.get('page', 1))
per_page = 10
inner_window=10
offset = (page) * 10
return page, per_page, offset, inner_window
def get_pagination(**kwargs):
kwargs.setdefault('record_name', 'repositories')
return Pagination(css_framework=get_css_framework(),
link_size=get_link_size(),
show_single_page=show_single_page_or_not(),
**kwargs
)

If you print out page you will probably see that it always returns 1 since you are trying to access the request outside of its scope, the search route function, and .get() is always returning 1.
Try something like this...
#search.route('/search')
def search():
...
page = int(request.args.get('page', 1))
per_page = 10
inner_window = 10
offset = page * 10
...
If you want to continue to use a separate function the you will need to pass the request object to it in order to access the values.
Hope that this helps!

I was able to solve this by switching from Flask-Paginate for the DB query to just SQLALchemy via Flask-Sqlalchemy.
I changed this to return desired results per page..
links = Item.query.paginate(page, 10, True)
More information is here (SQLAlchemy and paging) and on the `Flask-SqlAlchemy docs'.

I see that you opted to use Flask-SQLAlchemy instead, but hopefully this answer will help someone else:
You were almost there - you correctly captured page, per_page, and offset (there's actually a built-in for this, which you can import using from flask_paginate import get_page_args), but you also need to use the per_page and offset on your query. So in your example:
all_links = Item.query.all()
links_for_render_template = Item.query.limit(per_page).offset(offset)
Here's a full example as well:
from flask_paginate import Pagination, get_page_args
#search.route('/search')
def search():
page, per_page, offset = get_page_args()
all_links = Item.query.all()
links_for_render_template = Item.query.limit(per_page).offset(offset)
pagination = get_pagination(page=page, per_page=per_page, offset=offset, total=all_links.count(), record_name='links')
return render_template('search/searchPage.html', links=links_for_render_template, pagination=pagination)

python mechanize follow_link fails

I'm trying to access search results on the NCBI Images search page (http://www.ncbi.nlm.nih.gov/images) in a script. I want to feed it a search term, report on all of the results, and then move on to the next search term. To do this I need to get to results pages after the first page, so I'm trying to use python mechanize to do it:
import mechanize
browser=mechanize.Browser()
page1=browser.open('http://www.ncbi.nlm.nih.gov/images?term=drug')
a=browser.links(text_regex='Next')
nextlink=a.next()
page2=browser.follow_link(nextlink)
This just gives me back the first page of search results again (in variable page2). What am I doing wrong, and how can I get to that second page and beyond?

Unfortunately that page uses Javascript to POST 2459 bytes of form variables to the server, just to navigate to a subsequent page. Here are a few of the variables (I count 38 vars in total):
EntrezSystem2.PEntrez.ImagesDb.Images_SearchBar.Term=drug
EntrezSystem2.PEntrez.ImagesDb.Images_SearchBar.CurrDb=images
EntrezSystem2.PEntrez.ImagesDb.Images_ResultsPanel.Entrez_Pager.CurrPage=2
You'll need to construct a POST request to the server containing some or all of these variables. Luckily if you get it working for page 2 you can simply increment CurrPage and send another POST to get each subsequent page of results (no need to extract links).
Update - That site is a total pain-in-the-ass, but here is a POST-based scrape of the 2-N pages. Set MAX_PAGE to the highest page number + 1. The script will produce files like file_000003.html.
Note: Before you use it, you need to replace POSTDATA with the contents of this paste blob (it expires in 1 month). It's just the body a POST request as captured by Firebug, which I use to seed the correct params:
import cookielib
import json
import mechanize
import sys
import urllib
import urlparse
MAX_PAGE = 6
TERM = 'drug'
DEBUG = False
base_url = 'http://www.ncbi.nlm.nih.gov/images?term=' + TERM
browser = mechanize.Browser()
browser.set_handle_robots(False)
browser.set_handle_referer(True)
browser.set_debug_http(DEBUG)
browser.set_debug_responses(DEBUG)
cjar = cookielib.CookieJar()
browser.set_cookiejar(cjar)
# make first GET request. this will populate the cookie
res = browser.open(base_url)
def write(num, data):
with open('file_%06d.html' % num, 'wb') as out:
out.write(data)
def encode(kvs):
res = []
for key, vals in kvs.iteritems():
if isinstance(vals, list):
for v in vals:
res.append('%s=%s' % (key, urllib.quote(v)))
else:
res.append('%s=%s' % (key, urllib.quote(vals)))
return '&'.join(res)
write(1, res.read())
# set this var equal to the contents of this: http://pastebin.com/UfejW3G0
POSTDATA = '''<post data>'''
# parse the embedded json vars into POST parameters
PREFIX1 = 'EntrezSystem2.PEntrez.ImagesDb.'
PREFIX2 = 'EntrezSystem2.PEntrez.DbConnector.'
params = dict((k, v[0]) for k, v in urlparse.parse_qs(POSTDATA).iteritems())
base_url = 'http://www.ncbi.nlm.nih.gov/images'
for page in range(2, MAX_PAGE):
params[PREFIX1 + 'Images_ResultsPanel.Entrez_Pager.CurrPage'] = str(page)
params[PREFIX1 + 'Images_ResultsPanel.Entrez_Pager.cPage'] = [str(page-1)]*2
data = encode(params)
req = mechanize.Request(base_url, data)
cjar.add_cookie_header(req)
req.add_header('Content-Type', 'application/x-www-form-urlencoded')
req.add_header('Referer', base_url)
res = browser.open(req)
write(page, res.read())