Develop Reference

Python - web crawling / different result from same code? / requests, bs4 / M1

I learning python for web crawling, but i'm totally stuck.
Each time I run this codes, results change.
very rarely, it works but almost return empty list.
why does it happen? please let me know
from indeed import extract_indeed_pages, extract_indeed_jobs
last_indeed_page = extract_indeed_pages()
print(last_indeed_page)
indeed_jobs = extract_indeed_jobs(last_indeed_page)
print(indeed_jobs)
import requests
from bs4 import BeautifulSoup
LIMIT = 50
URL = f"https://kr.indeed.com/jobs?q=React&l=%EC%84%9C%EC%9A%B8&radius=100&jt=fulltime&limit={LIMIT}"
def extract_indeed_pages():
result = requests.get(URL)
soup = BeautifulSoup(result.text, "html.parser")
pagination = soup.find("div", {"class": "pagination"})
links = pagination.find_all('a')
pages = []
for link in links[:-1]:
pages.append(int(link.string))
max_page = pages[-1]
return max_page
def extract_indeed_jobs(last_page):
jobs = []
result = requests.get(f"{URL}&start={0*LIMIT}")
soup = BeautifulSoup(result.text, "html.parser")
results = soup.find_all("h2", {"class": "jobTitle"})
jobs.append(results)
return jobs

This happens because of the javascript on the source code. You can view the web page by pressing the ctrl + u buttons on your pc.

webscraping bus stops with beautifulsoup

I am trying to web scrape bus stop names for a given line, here is an example page for line 212 https://www.m2.rozkladzik.pl/warszawa/rozklad_jazdy.html?l=212. I want to have as an output two lists, one with bus stop names in one direction and the other list with another direction. (It's clearly seen on the web page). I managed to get all names in one list with
import requests
from bs4 import BeautifulSoup
def download_bus_schedule(bus_number):
URL = "http://www.m2.rozkladzik.pl/warszawa/rozklad_jazdy.html?l=" + bus_number
r = requests.get(URL)
soup = BeautifulSoup(r.content,
'html5lib')
print(soup.prettify())
all_bus_stops = []
table = soup.find_all('a')
for element in table:
if element.get_text() in all_bus_stops:
continue
else:
all_bus_stops.append(element.get_text())
return all_bus_stops
print(download_bus_schedule('212'))
I guess the solution would be to somehow divide the soup into two parts.

You can use the bs4.element.Tag.findAll method:
import requests
from bs4 import BeautifulSoup
def download_bus_schedule(bus_number):
all_bus_stops = []
URL = "http://www.m2.rozkladzik.pl/warszawa/rozklad_jazdy.html?l=" + bus_number
r = requests.get(URL)
soup = BeautifulSoup(r.content, 'html.parser')
for s in soup.select(".holo-list"):
bus_stops = []
for f in s.findAll("li"):
if f.text not in bus_stops:
bus_stops.append(f.text)
all_bus_stops.append(bus_stops)
return all_bus_stops
print(download_bus_schedule('212'))
Output:
[['Pl.Hallera', 'Pl.Hallera', 'Darwina', 'Namysłowska', 'Rondo Żaba', 'Rogowska', 'Kołowa', 'Dks Targówek', 'Metro Targówek Mieszkaniowy', 'Myszkowska', 'Handlowa', 'Metro Trocka', 'Bieżuńska', 'Jórskiego', 'Łokietka', 'Samarytanka', 'Rolanda', 'Żuromińska', 'Targówek-Ratusz', 'Św.Wincentego', 'Malborska', 'Ch Targówek'],
['Ch Targówek', 'Ch Targówek', 'Malborska', 'Św.Wincentego', 'Targówek-Ratusz', 'Żuromińska', 'Gilarska', 'Rolanda', 'Samarytanka', 'Łokietka', 'Jórskiego', 'Bieżuńska', 'Metro Trocka', 'Metro Trocka', 'Metro Trocka', 'Handlowa', 'Myszkowska', 'Metro Targówek Mieszkaniowy', 'Dks Targówek', 'Kołowa', 'Rogowska', 'Rondo Żaba', '11 Listopada', 'Bródnowska', 'Szymanowskiego', 'Pl.Hallera', 'Pl.Hallera']]

import requests
from bs4 import BeautifulSoup
def download_bus_schedule(bus_number):
URL = "http://www.m2.rozkladzik.pl/warszawa/rozklad_jazdy.html?l=" + bus_number
r = requests.get(URL)
soup = BeautifulSoup(r.content,
'html5lib')
bus_stops_1 = []
bus_stops_2 = []
directions = soup.find_all("ul", {"class":"holo-list"})
for stop in directions[0].find_all("a"):
if stop not in bus_stops_1:
bus_stops_1.append(stop.text.strip())
for stop in directions[1].find_all("a"):
if stop not in bus_stops_2:
bus_stops_2.append(stop.text.strip())
all_bus_stops = (bus_stops_1, bus_stops_2)
return all_bus_stops
print(download_bus_schedule('212')[0])
print(download_bus_schedule('212')[1])

I may have misunderstood as I do not know Polish but see if this helps.
from bs4 import BeautifulSoup
import requests
url = 'https://www.m2.rozkladzik.pl/warszawa/rozklad_jazdy.html?l=212'
resp = requests.get(url)
soup = BeautifulSoup(resp.content, "html.parser")
d = {}
for h2 in soup.select('h2.holo-divider'):
d[h2.text] = []
ul = h2.next_sibling
for li in ul.select('li'):
if li.a.text not in d[h2.text]:
d[h2.text].append(li.a.text)
from pprint import pprint
pprint(d)

As all stops are encapsulated in the next un-ordered list, you could use the find_next function of bs4.
e.g.
URL = f"http://www.m2.rozkladzik.pl/warszawa/rozklad_jazdy.html?l={bus_number}"
r = requests.get(URL)
soup = BeautifulSoup(r.content,
'html5lib')
directions = ["Ch Targówek","Pl.Hallera"]
result = {}
for direction in directions:
header = soup.find(text=direction)
list = header.find_next("ul")
stops_names = [stop.get_text() for stop in list]
result[direction] = stops_names
return result
Plus you might want to use f-string to format your strings as it improves reading and is less error prone.

Running a function in parallel and saving the return in a list using python

I have a function that scrapes href link form a particular page and returns the result. I want to call this function in a parallel way to save time. I have visited this problem Running same function for multiple files in parallel in python
But the challenge is that I need to save the return element in a list. How can I do that? Here is my code snippet.
url = "https://www.programmableweb.com/category/all/apis"
response = requests.get(url)
data = response.text
soup = BeautifulSoup(data,'html.parser')
#function to scrape individual pages
def scrap_api_url(i):
print(i)
page_url = "https://www.programmableweb.com" + mid_url + '=' + str(i)
response = requests.get(page_url)
data = response.text
soup = BeautifulSoup(data, 'html.parser')
all_api = soup.find_all('tr', class_ = re.compile('^(even|odd)$'))
return all_api
url_tag = soup.find('a',{'title' : 'Go to next page'})
mid_url = url_tag.get('href').split('=')[0]
threads=[]
#calling functions
if __name__ == '__main__':
inputs = [i for i in range(851)]
for item in inputs:
print('Thread Started :: ', item)
t = threading.Thread(target = scrap_api_url, args=(item,))
threads.append(t)
t.start()
h = []
for t in threads:
h.append(t.join())

You can use the ThreadPoolExecutor map method:
import re
from concurrent.futures import ThreadPoolExecutor
import requests
from bs4 import BeautifulSoup
def main():
url = "https://www.programmableweb.com/category/all/apis"
response = requests.get(url)
data = response.text
soup = BeautifulSoup(data,'html.parser')
url_tag = soup.find('a',{'title' : 'Go to next page'})
mid_url = url_tag.get('href').split('=')[0]
# function to scrape individual pages
def scrap_api_url(i):
print(i)
page_url = "https://www.programmableweb.com" + mid_url + '=' + str(i)
response = requests.get(page_url)
data = response.text
soup = BeautifulSoup(data, 'html.parser')
all_api = soup.find_all('tr', class_=re.compile('^(even|odd)$'))
return all_api
inputs = [i for i in range(851)]
with ThreadPoolExecutor() as executor:
future_results = executor.map(scrap_api_url, inputs)
results = [result for result in future_results]
print(results)
#calling functions
if __name__ == '__main__':
main()

How to Iterate over pages in Ebay

I am building a scraper for Ebay. I am trying to figure out a way to manipulate the page number portion of the Ebay url to go to the next page until there are no more pages (If you were on page 2 the page number portion would look like "_pgn=2"). I noticed that if you put any number greater than the max number of pages a listing has, the page will reload to the last page, not give like a page doesn't exist error. (If a listing has 5 pages, then the last listing' page number url portion of _pgn=5 would rout to the same page if the page number url portion was _pgn=100). How can I implement a way to start at page one, get the html soup of the page, get the all relevant data I want from the soup, then load up the next page with the new page number and start the process again until there are not any new pages to scrape? I tried to get the number of results a listing has by using selenium xpath and math.ceil the quotient of number of results and 50 (default number of max listings per page) and use that quotient as my max_page, but I get errors saying the element doesn't exist even though it does. self.driver.findxpath('xpath').text. That 243 is what I am trying to get with the xpath.
class EbayScraper(object):
def __init__(self, item, buying_type):
self.base_url = "https://www.ebay.com/sch/i.html?_nkw="
self.driver = webdriver.Chrome(r"chromedriver.exe")
self.item = item
self.buying_type = buying_type + "=1"
self.url_seperator = "&_sop=12&rt=nc&LH_"
self.url_seperator2 = "&_pgn="
self.page_num = "1"
def getPageUrl(self):
if self.buying_type == "Buy It Now=1":
self.buying_type = "BIN=1"
self.item = self.item.replace(" ", "+")
url = self.base_url + self.item + self.url_seperator + self.buying_type + self.url_seperator2 + self.page_num
response = requests.get(url)
soup = BeautifulSoup(response.text, 'html.parser')
return soup
def getInfo(self, soup):
for listing in soup.find_all("li", {"class": "s-item"}):
raw = listing.find_all("a", {"class": "s-item__link"})
if raw:
raw_price = listing.find_all("span", {"class": "s-item__price"})[0]
raw_title = listing.find_all("h3", {"class": "s-item__title"})[0]
raw_link = listing.find_all("a", {"class": "s-item__link"})[0]
raw_condition = listing.find_all("span", {"class": "SECONDARY_INFO"})[0]
condition = raw_condition.text
price = float(raw_price.text[1:])
title = raw_title.text
link = raw_link['href']
print(title)
print(condition)
print(price)
if self.buying_type != "BIN=1":
raw_time_left = listing.find_all("span", {"class": "s-item__time-left"})[0]
time_left = raw_time_left.text[:-4]
print(time_left)
print(link)
print('\n')
if __name__ == '__main__':
item = input("Item: ")
buying_type = input("Buying Type (e.g, 'Buy It Now' or 'Auction'): ")
instance = EbayScraper(item, buying_type)
page = instance.getPageUrl()
instance.getInfo(page)

if you want to iterate all pages and gather all results then your script needs to check if there is a next page after you visit the page
import requests
from bs4 import BeautifulSoup
class EbayScraper(object):
def __init__(self, item, buying_type):
...
self.currentPage = 1
def get_url(self, page=1):
if self.buying_type == "Buy It Now=1":
self.buying_type = "BIN=1"
self.item = self.item.replace(" ", "+")
# _ipg=200 means that expect a 200 items per page
return '{}{}{}{}{}{}&_ipg=200'.format(
self.base_url, self.item, self.url_seperator, self.buying_type,
self.url_seperator2, page
)
def page_has_next(self, soup):
container = soup.find('ol', 'x-pagination__ol')
currentPage = container.find('li', 'x-pagination__li--selected')
next_sibling = currentPage.next_sibling
if next_sibling is None:
print(container)
return next_sibling is not None
def iterate_page(self):
# this will loop if there are more pages otherwise end
while True:
page = instance.getPageUrl(self.currentPage)
instance.getInfo(page)
if self.page_has_next(page) is False:
break
else:
self.currentPage += 1
def getPageUrl(self, pageNum):
url = self.get_url(pageNum)
print('page: ', url)
response = requests.get(url)
soup = BeautifulSoup(response.text, 'html.parser')
return soup
def getInfo(self, soup):
...
if __name__ == '__main__':
item = input("Item: ")
buying_type = input("Buying Type (e.g, 'Buy It Now' or 'Auction'): ")
instance = EbayScraper(item, buying_type)
instance.iterate_page()
the important functions here are page_has_next and iterate_page
page_has_next - a function that check if the pagination of the page has another li element next to the selected page. e.g < 1 2 3 > if we are on page 1 then it checks if there is 2 next -> something like this
iterate_page - a function that loop until there is no page_next
also note that you don't need selenium for this unless you need to mimic user clicks or need a browser to navigate.

How can I crawl all the page?

I am trying to crawl sites's text. But It's only crawling 12 articles.
I don't know why does it do like that. and I wondering If I wanna crawl other pages, What should I do?
import requests
from bs4 import BeautifulSoup
x = int(input("start page:"))
while x < int(input("end page:")):
x = x + 1
url = "https://www.mmtimes.com/national-news.html?page=" + str(x)
result = requests.get(url)
bs_obj = BeautifulSoup(result.content, "html.parser")
content = bs_obj.find("div", {"class": "msp-three-col"})
read_more = content.findAll("div", {"class": "read-more"})
for item in read_more:
atag = item.find('a')
link = "https://www.mmtimes.com" + atag["href"]
linkResult = requests.get(link)
subpage = BeautifulSoup(linkResult.content, "html.parser")
fnresult = subpage.find("div", {"class": "field-item even"})
print(fnresult.text)
print("Total "+str(len(read_more))+" articles"))

Check out the below code, I have made some changes. this will result the required output.
import requests
from bs4 import BeautifulSoup
x = int(input("start page:"))
y = input("end page:")
article_count = 0
while x <= int(y):
url = "https://www.mmtimes.com/national-news.html?page=" + str(x)
result = requests.get(url)
bs_obj = BeautifulSoup(result.content, "html.parser")
content = bs_obj.find("div", {"class": "msp-three-col"})
read_more = content.findAll("div", {"class": "read-more"})
for item in read_more:
atag = item.find('a')
link = "https://www.mmtimes.com" + atag["href"]
linkResult = requests.get(link)
subpage = BeautifulSoup(linkResult.content, "html.parser")
fnresult = subpage.find("div", {"class": "field-item even"})
print(fnresult.text)
article_count += len(read_more)
print("Total "+str(article_count)+" articles")
x += 1