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how to scrape data on a website with view more with beautifulsoup

i am trying to scrape news from reuters but there is a click to view more at the bottom on the website. I could not know how to load the hidden results by using beautiful soup.
from bs4 import BeautifulSoup
import urllib.request
def scrape_reuters_news(ticker):
url = "https://www.reuters.com/search/news?sortBy=relevance&dateRange=pastWeek&blob="+ticker
scraped_data = urllib.request.urlopen(url)
scraped_data = scraped_data.read()
parsed_articles = BeautifulSoup(scraped_data, 'lxml')
links = parsed_articles.find_all("h3")
articles = []
titles = []
title_class = "Text__text___3eVx1j Text__dark-grey___AS2I_p Text__medium___1ocDap Text__heading_2___sUlNJP Heading__base___1dDlXY Heading__heading_2___3f_bIW ArticleHeader__heading___3ibi0Q"
for link in links:
paragraphs = ""
url = "https://www.reuters.com/"+str(link)[41:63]
scraped_data = urllib.request.urlopen(url)
scraped_data = scraped_data.read()
parsed_article = BeautifulSoup(scraped_data, 'lxml')
article = parsed_article.find_all("p")
title = parsed_article.select("h1", {"class": title_class})
titles.append(title[0].text.strip())
for paragraph in article:
paragraphs += paragraph.text + " "
articles.append(paragraphs)
return titles, articles
# edit
ticker = "apple"
news = scrape_reuters_news(ticker)

When you click the load more a callback is issued that you can find in the network tab. If you grab the number of results from the search page, you can add this into the callback to get all results in one go. I then use regex to extract the id to reconstruct each detail page url and the title (headline)
You would then visit each link to get the paragraph info.
Please note:
There is some de-duplication work to do. There exist different ids which lead to same content. So perhaps exclude based on title?
You may need to consider whether any pre-processing of ticker needs to happen e.g. convert to lowercase, replace spaces with "-". I don't know all your use cases.
from bs4 import BeautifulSoup as bs
import requests, re
ticker = 'apple'
with requests.Session() as s:
r = s.get(f'https://www.reuters.com/search/news?sortBy=relevance&dateRange=pastWeek&blob={ticker}')
soup = bs(r.content, 'lxml')
num_results = soup.select_one('.search-result-count-num').text
r = s.get(f'https://www.reuters.com/assets/searchArticleLoadMoreJson?blob={ticker}&bigOrSmall=big&articleWithBlog=true&sortBy=relevance&dateRange=pastWeek&numResultsToShow={num_results}&pn=&callback=addMoreNewsResults')
p = re.compile(r'id: "(.*?)"')
p2 = re.compile(r'headline: "(.*?)"')
links = [f'https://www.reuters.com/article/id{i}' for i in p.findall(r.text)]
headlines = [bs(i, 'lxml').get_text() for i in p2.findall(r.text)]
print(len(links), len(headlines))
From the detail pages you can get the paragraphs with
paras = ' '.join([i.get_text() for i in soup.select('[data-testid*=paragraph-]')])

python crawling beautifulsoup how to crawl several pages?

Please Help.
I want to get all the company names of each pages and they have 12 pages.
http://www.saramin.co.kr/zf_user/jobs/company-labs/list/page/1
http://www.saramin.co.kr/zf_user/jobs/company-labs/list/page/2
-- this website only changes the number.
So Here is my code so far.
Can I get just the title (company name) of 12 pages?
Thank you in advance.
from bs4 import BeautifulSoup
import requests
maximum = 0
page = 1
URL = 'http://www.saramin.co.kr/zf_user/jobs/company-labs/list/page/1'
response = requests.get(URL)
source = response.text
soup = BeautifulSoup(source, 'html.parser')
whole_source = ""
for page_number in range(1, maximum+1):
URL = 'http://www.saramin.co.kr/zf_user/jobs/company-labs/list/page/' + str(page_number)
response = requests.get(URL)
whole_source = whole_source + response.text
soup = BeautifulSoup(whole_source, 'html.parser')
find_company = soup.select("#content > div.wrap_analysis_data > div.public_con_box.public_list_wrap > ul > li:nth-child(13) > div > strong")
for company in find_company:
print(company.text)
---------Output of one page
---------page source :)

So, you want to remove all the headers and get only the string of the company name?
Basically, you can use the soup.findAll to find the list of company in the format like this:
<strong class="company"><span>중소기업진흥공단</span></strong>
Then you use the .find function to extract information from the <span> tag:
<span>중소기업진흥공단</span>
After that, you use .contents function to get the string from the <span> tag:
'중소기업진흥공단'
So you write a loop to do the same for each page, and make a list called company_list to store the results from each page and append them together.
Here's the code:
from bs4 import BeautifulSoup
import requests
maximum = 12
company_list = [] # List for result storing
for page_number in range(1, maximum+1):
URL = 'http://www.saramin.co.kr/zf_user/jobs/company-labs/list/page/{}'.format(page_number)
response = requests.get(URL)
print(page_number)
whole_source = response.text
soup = BeautifulSoup(whole_source, 'html.parser')
for entry in soup.findAll('strong', attrs={'class': 'company'}): # Finding all company names in the page
company_list.append(entry.find('span').contents[0]) # Extracting name from the result
The company_list will give you all the company names you want

I figured it out eventually. Thank you for your answer though!
image : code captured in jupyter notebook
Here is my final code.
from urllib.request import urlopen
from bs4 import BeautifulSoup
company_list=[]
for n in range(12):
url = 'http://www.saramin.co.kr/zf_user/jobs/company-labs/list/page/{}'.format(n+1)
webpage = urlopen(url)
source = BeautifulSoup(webpage,'html.parser',from_encoding='utf-8')
companys = source.findAll('strong',{'class':'company'})
for company in companys:
company_list.append(company.get_text().strip().replace('\n','').replace('\t','').replace('\r',''))
file = open('company_name1.txt','w',encoding='utf-8')
for company in company_list:
file.write(company+'\n')
file.close()

Trouble parsing product names out of some links with different depth

I've written a script in python to reach the target page where each category has their avaiable item names in a website. My below script can get the product names from most of the links (generated through roving category links and then subcategory links).
The script can parse sub-category links revealed upon clicking + sign located right next to each category which are visible in the below image and then parse all the product names from the target page. This is one of such target pages.
However, few of the links do not have the same depth as other links. For example this link and this one are different from usual links like this one.
How can I get all the product names from all the links irrespective of their different depth?
This is what I've tried so far:
import requests
from urllib.parse import urljoin
from bs4 import BeautifulSoup
link = "https://www.courts.com.sg/"
res = requests.get(link)
soup = BeautifulSoup(res.text,"lxml")
for item in soup.select(".nav-dropdown li a"):
if "#" in item.get("href"):continue #kick out invalid links
newlink = urljoin(link,item.get("href"))
req = requests.get(newlink)
sauce = BeautifulSoup(req.text,"lxml")
for elem in sauce.select(".product-item-info .product-item-link"):
print(elem.get_text(strip=True))
How to find trget links:

The site has six main product categories. Products that belong to a subcategory can also be found in a main category (for example the products in /furniture/furniture/tables can also be found in /furniture), so you only have to collect products from the main categories. You could get the categories links from the main page, but it'd be easier to use the sitemap.
url = 'https://www.courts.com.sg/sitemap/'
r = requests.get(url)
soup = BeautifulSoup(r.text, 'html.parser')
cats = soup.select('li.level-0.category > a')[:6]
links = [i['href'] for i in cats]
As you've mentioned there are some links that have differend structure, like this one: /televisions. But, if you click the View All Products link on that page you will be redirected to /tv-entertainment/vision/television. So, you can get all the /televisions rpoducts from /tv-entertainment. Similarly, the products in links to brands can be found in the main categories. For example, the /asus products can be found in /computing-mobile and other categories.
The code below collects products from all the main categories, so it should collect all the products on the site.
from bs4 import BeautifulSoup
import requests
url = 'https://www.courts.com.sg/sitemap/'
r = requests.get(url)
soup = BeautifulSoup(r.text, 'html.parser')
cats = soup.select('li.level-0.category > a')[:6]
links = [i['href'] for i in cats]
products = []
for link in links:
link += '?product_list_limit=24'
while link:
r = requests.get(link)
soup = BeautifulSoup(r.text, 'html.parser')
link = (soup.select_one('a.action.next') or {}).get('href')
for elem in soup.select(".product-item-info .product-item-link"):
product = elem.get_text(strip=True)
products += [product]
print(product)
I've increased the number of products per page to 24, but still this code takes a lot of time, as it collects products from all main categories and their pagination links. However, we could make it much faster with the use of threads.
from bs4 import BeautifulSoup
import requests
from threading import Thread, Lock
from urllib.parse import urlparse, parse_qs
lock = Lock()
threads = 10
products = []
def get_products(link, products):
soup = BeautifulSoup(requests.get(link).text, 'html.parser')
tags = soup.select(".product-item-info .product-item-link")
with lock:
products += [tag.get_text(strip=True) for tag in tags]
print('page:', link, 'items:', len(tags))
url = 'https://www.courts.com.sg/sitemap/'
soup = BeautifulSoup(requests.get(url).text, 'html.parser')
cats = soup.select('li.level-0.category > a')[:6]
links = [i['href'] for i in cats]
for link in links:
link += '?product_list_limit=24'
soup = BeautifulSoup(requests.get(link).text, 'html.parser')
last_page = soup.select_one('a.page.last')['href']
last_page = int(parse_qs(urlparse(last_page).query)['p'][0])
threads_list = []
for i in range(1, last_page + 1):
page = '{}&p={}'.format(link, i)
thread = Thread(target=get_products, args=(page, products))
thread.start()
threads_list += [thread]
if i % threads == 0 or i == last_page:
for t in threads_list:
t.join()
print(len(products))
print('\n'.join(products))
This code collects 18,466 products from 773 pages in about 5 minutes. I'm using 10 threads because I don't want to stress the server too much, but you could use more (most servers can handle 20 threads easily).

I would recommend starting your scrape from the pages sitemap
Found here
If they were to add products, it's likely to show up here as well.

Since your main issue is finding the links, here is a generator that will find all of the category and sub-category links using the sitemap krflol pointed out in his solution:
from bs4 import BeautifulSoup
import requests
def category_urls():
response = requests.get('https://www.courts.com.sg/sitemap')
html_soup = BeautifulSoup(response.text, features='html.parser')
categories_sitemap = html_soup.find(attrs={'class': 'xsitemap-categories'})
for category_a_tag in categories_sitemap.find_all('a'):
yield category_a_tag.attrs['href']
And to find the product names, simply scrape each of the yielded category_urls.

I saw the website for parsing and found that all the products are available at the bottom left side of the main page https://www.courts.com.sg/ .After clicking one of these we goes to advertisement front page of a particular category. Where we have to go in click All Products for getting it.
Following is the code as whole:
import requests
from bs4 import BeautifulSoup
def parser():
parsing_list = []
url = 'https://www.courts.com.sg/'
source_code = requests.get(url)
plain_text = source_code.text
soup = BeautifulSoup(plain_text, "html.parser")
ul = soup.find('footer',{'class':'page-footer'}).find('ul')
for l in ul.find_all('li'):
nextlink = url + l.find('a').get('href')
response = requests.get(nextlink)
inner_soup = BeautifulSoup(response.text, "html.parser")
parsing_list.append(url + inner_soup.find('div',{'class':'category-static-links ng-scope'}).find('a').get('href'))
return parsing_list
This function will return list of all products of all categories which your code didn't scrape from it.

Webscraping multiple pages from a site gives just the home page contents?

I am trying to scrape the pages from the site https://www.rithmschool.com/blog. Though the contents of the 1st page are getting scraped, the problem with my code is - the same content is being scraped even for all the other pages of the site. Below is the code.
Can anyone please help me fix the code.
My code
import requests
from bs4 import BeautifulSoup
from csv import writer
html = requests.get('https://www.rithmschool.com/blog')
soup = BeautifulSoup(html.text, 'html.parser')
articles = soup.find_all('article')
with open('scraped_rithm.csv', 'w') as f:
data = writer(f)
data.writerow(['Title','Link','Date'])
for article in articles:
title = article.find('a').get_text()
link = article.find('a')['href']
date = article.find('time')['datetime']
data.writerow([title,link,date])
spans = soup.find_all('span', {'class' : 'page'})
for span in spans:
if span.find('a'):
urls = ((span.find('a')['href'])).split(',')
for url in urls:
nw_urls = (f"https://www.rithmschool.com{url}")
print(nw_urls)
nw_response = requests.get(nw_urls)
nw_soup = BeautifulSoup(nw_response.text,'html.parser')
articles = soup.find_all('article')
for article in articles:
title = article.find('a').get_text()
link = article.find('a')['href']
date = article.find('time')['datetime']
data.writerow([title,link,date])

You are still using the old soup object after creating a new nw_soup object
Try changing this from
nw_soup = BeautifulSoup(nw_response.text,'html.parser')
articles = soup.find_all('article')
to
nw_soup = BeautifulSoup(nw_response.text,'html.parser')
articles = nw_soup.find_all('article')

How to make my crawler parse data from start page

I've written some code in python to grab details from a torrent site. However, when I run my code I found the results as I expected. The only problem with this crawler is that it skips the content of first page [as the pagination urls start from 2] which I can't fix. Any help on this will be highly appreciable.
import requests
from lxml import html
page_link = "https://yts.ag/browse-movies"
b_link = "https://yts.ag"
def get_links(main_link):
response = requests.get(main_link).text
tree = html.fromstring(response)
for item in tree.cssselect('ul.tsc_pagination a'):
if "page" in item.attrib["href"]:
movie_details(b_link + item.attrib["href"])
def movie_details(link):
response = requests.get(link).text
tree = html.fromstring(response)
for titles in tree.cssselect("div.browse-movie-wrap"):
title = titles.cssselect('div.browse-movie-bottom a.browse-movie-title')[0].text
link = titles.cssselect('div.browse-movie-year')[0].text
rating= titles.cssselect('figcaption.hidden-xs h4.rating')[0].text
genre = titles.cssselect('figcaption.hidden-xs h4')[0].text
genre1 = titles.cssselect('figcaption.hidden-xs h4')[1].text
print(title, link, rating, genre, genre1)
get_links(page_link)

Why not just call the movie_details() function on the main_link before the loop ?
def get_links(main_link):
response = requests.get(main_link).text
tree = html.fromstring(response)
movie_details(main_link)
for item in tree.cssselect('ul.tsc_pagination a'):
if "page" in item.attrib["href"]:
movie_details(b_link + item.attrib["href"])