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Pulling p tags from multiple URLs

I've struggled on this for days and not sure what the issue could be - basically, I'm trying to extract the profile box data (picture below) of each link -- going through inspector, I thought I could pull the p tags and do so.
I'm new to this and trying to understand, but here's what I have thus far:
-- a code that (somewhat) succesfully pulls the info for ONE link:
import requests
from bs4 import BeautifulSoup
# getting html
url = 'https://basketball.realgm.com/player/Darius-Adams/Summary/28720'
req = requests.get(url)
soup = BeautifulSoup(req.text, 'html.parser')
container = soup.find('div', attrs={'class', 'main-container'})
playerinfo = container.find_all('p')
print(playerinfo)
I then also have a code that pulls all of the HREF tags from multiple links:
from bs4 import BeautifulSoup
import requests
def get_links(url):
links = []
website = requests.get(url)
website_text = website.text
soup = BeautifulSoup(website_text)
for link in soup.find_all('a'):
links.append(link.get('href'))
for link in links:
print(link)
print(len(links))
get_links('https://basketball.realgm.com/dleague/players/2022')
get_links('https://basketball.realgm.com/dleague/players/2021')
get_links('https://basketball.realgm.com/dleague/players/2020')
So basically, my goal is to combine these two, and get one code that will pull all of the P tags from multiple URLs. I've been trying to do it, and I'm really not sure at all why this isn't working here:
from bs4 import BeautifulSoup
import requests
def get_profile(url):
profiles = []
req = requests.get(url)
soup = BeautifulSoup(req.text, 'html.parser')
container = soup.find('div', attrs={'class', 'main-container'})
for profile in container.find_all('a'):
profiles.append(profile.get('p'))
for profile in profiles:
print(profile)
get_profile('https://basketball.realgm.com/player/Darius-Adams/Summary/28720')
get_profile('https://basketball.realgm.com/player/Marial-Shayok/Summary/26697')
Again, I'm really new to web scraping with Python but any advice would be greatly appreciated. Ultimately, my end goal is to have a tool that can scrape this data in a clean way all at once.
(Player name, Current Team, Born, Birthplace, etc).. maybe I'm doing it entirely wrong but any guidance is welcome!

You need to combine your two scripts together and make requests for each player. Try the following approach. This searches for <td> tags that have the data-td=Player attribute:
import requests
from bs4 import BeautifulSoup
def get_links(url):
data = []
req_url = requests.get(url)
soup = BeautifulSoup(req_url.content, "html.parser")
for td in soup.find_all('td', {'data-th' : 'Player'}):
a_tag = td.a
name = a_tag.text
player_url = a_tag['href']
print(f"Getting {name}")
req_player_url = requests.get(f"https://basketball.realgm.com{player_url}")
soup_player = BeautifulSoup(req_player_url.content, "html.parser")
div_profile_box = soup_player.find("div", class_="profile-box")
row = {"Name" : name, "URL" : player_url}
for p in div_profile_box.find_all("p"):
try:
key, value = p.get_text(strip=True).split(':', 1)
row[key.strip()] = value.strip()
except: # not all entries have values
pass
data.append(row)
return data
urls = [
'https://basketball.realgm.com/dleague/players/2022',
'https://basketball.realgm.com/dleague/players/2021',
'https://basketball.realgm.com/dleague/players/2020',
]
for url in urls:
print(f"Getting: {url}")
data = get_links(url)
for entry in data:
print(entry)

Loop through Hrefs in Python Using BeautifulSoup

So I am trying to extract all the article URLs from this link.
However I only get ['https://mn.usembassy.gov/mn/2020-naadam-mn/', 'https://mn.usembassy.gov/mn/06272020-presidential-proclamation-mn/', 'https://mn.usembassy.gov/mn/pr-060320-mn/', 'https://mn.usembassy.gov/mn/dv-2021-status-check-mn/', 'https://mn.usembassy.gov/mn/pr-050120-mn/']
below is my code currently.
The website has 52 pages and I am trying to get all the URLs, why does it only give me a few URLs instead of all the URLS?
import requests
from bs4 import BeautifulSoup
url = 'https://mn.usembassy.gov/mn/news-events-mn/'
reqs = requests.get(url)
soup = BeautifulSoup(reqs.text, 'lxml')
urls = []
for h in soup.find_all('h2'):
a = h.find('a')
urls.append(a.attrs['href'])
print(urls)

The page contains only 5 links to articles, you need to go to next page to load next 5 links. This script will get all links from the page:
import requests
from bs4 import BeautifulSoup
url = 'https://mn.usembassy.gov/mn/news-events-mn/page/{page}/'
urls = []
for page in range(1, 53):
soup = BeautifulSoup(requests.get(url.format(page=page)).content, 'html.parser')
for h in soup.find_all('h2'):
a = h.find('a')
print(a['href'])
urls.append(a.attrs['href'])
from pprint import pprint
pprint(urls)
Prints:
https://mn.usembassy.gov/mn/2020-naadam-mn/
https://mn.usembassy.gov/mn/06272020-presidential-proclamation-mn/
https://mn.usembassy.gov/mn/pr-060320-mn/
https://mn.usembassy.gov/mn/dv-2021-status-check-mn/
https://mn.usembassy.gov/mn/pr-050120-mn/
https://mn.usembassy.gov/mn/pr-042320-mca-website-mn/
https://mn.usembassy.gov/mn/2020-pr-us-mongolia-cpc-mn/
https://mn.usembassy.gov/mn/lead-2020-in-country-mn/
https://mn.usembassy.gov/mn/press-release-usaid-mar-24-2020-mn/
https://mn.usembassy.gov/mn/event-suspension-of-nonimmigrant-and-immigrant-visa-services-due-to-local-covid-19-related-preventative-measures-and-limited-staffing-mn/
https://mn.usembassy.gov/mn/2020-best-program-pr-mn/
https://mn.usembassy.gov/mn/2020-ncov-info-for-visa-mn/
...and so on.

Get links from a site's homepage using python

I want to write a script to get a home page's links to social media (twitter / facebook mostly), and I'm completely stuck since I am fairly new to Python.
The task I want to accomplish is to parse the website, find the social media links, and save it in a new data frame where each column would contain the original URL, the twitter link, and the facebook link. Here's what I have so far of this code for the new york times website:
from bs4 import BeautifulSoup
import requests
url = "http://www.nytimes.com"
r = requests.get(url)
sm_sites = ['twitter.com','facebook.com']
soup = BeautifulSoup(r.content, 'html5lib')
all_links = soup.find_all('a', href = True)
for site in sm_sites:
if all(site in sm_sites for link in all_links):
print(site)
else:
print('no link')
I'm having some problems understanding what the loop is doing, or how to make it work for what I need it to. I also had tried to store the site instead of doing print(site) but that was not working... So I figured I'd ask for help. Before asking, I went through a bunch of responses here but none could get me to do what I needed to do.

the way this code works, you already have your links. Your homepage link is the starting url, so http://www.nytimes.com.
And you have the social media urls sm_sites = ['twitter.com','facebook.com'], all you're doing is confirming they exist on the main page. If you want to save the list of confirmed social media urls, then append them to a list
Here is one way to get the social media links off a page
import requests
from bs4 import BeautifulSoup
url = "https://stackoverflow.com/questions/tagged/python"
r = requests.get(url)
sm_sites = ['twitter.com','facebook.com']
sm_sites_present = []
soup = BeautifulSoup(r.content, 'html5lib')
all_links = soup.find_all('a', href = True)
for sm_site in sm_sites:
for link in all_links:
if sm_site in link.attrs['href']:
sm_sites_present.append(link.attrs['href'])
print(sm_sites_present)
output:
['https://twitter.com/stackoverflow', 'https://www.facebook.com/officialstackoverflow/']
Update
for a df of urls
import requests
import pandas as pd
from bs4 import BeautifulSoup
from IPython.display import display
urls = [
"https://stackoverflow.com/questions/tagged/python",
"https://www.nytimes.com/",
"https://en.wikipedia.org/"
]
sm_sites = ['twitter.com','facebook.com']
sm_sites_present = []
columns = ['url'] + sm_sites
df = pd.DataFrame(data={'url' : urls}, columns=columns)
def get_sm(row):
r = requests.get(row['url'])
output = pd.Series()
soup = BeautifulSoup(r.content, 'html5lib')
all_links = soup.find_all('a', href = True)
for sm_site in sm_sites:
for link in all_links:
if sm_site in link.attrs['href']:
output[sm_site] = link.attrs['href']
return output
sm_columns = df.apply(get_sm, axis=1)
df.update(sm_columns)
df.fillna(value='no link')
output

This will do what you want with regards to adding it to a DataFrame. You can iterate through a list of websites (urlsToSearch), adding a row to the dataframe for each one containing the base website, all facebook links, and all twitter links.
from bs4 import BeautifulSoup
import requests
import pandas as pd
df = pd.DataFrame(columns=["Website", "Facebook", "Twitter"])
urlsToSearch = ["http://www.nytimes.com","http://www.businessinsider.com/"]
for url in urlsToSearch:
r = requests.get(url)
tw_links = []
fb_links = []
soup = BeautifulSoup(r.text, 'html.parser')
all_links = [link['href'] for link in soup.find_all('a', href = True)] #only get href
for link in all_links:
if "twitter.com" in link:
tw_links.append(link)
elif "facebook.com" in link:
fb_links.append(link)
df.loc[df.shape[0]] = [url,fb_links,tw_links] #Add row to end of df

how do I create a list from a sitemap.xml file to extract the url in python?

I need to create a code to extract a word from one scrape of images.
I'll explain, from a page sitemap.xml ,my code must try in every link present in this xml file, found insiede each link if there a specific word, inside an image link.
the sitemap is adidas = http://www.adidas.it/on/demandware.static/-/Sites-adidas-IT-Library/it_IT/v/sitemap/product/adidas-IT-it-it-product.xml
this is the code i created for search the image contains the word "ZOOM" :
import requests
from bs4 import BeautifulSoup
html = requests.get(
'http://www.adidas.it/scarpe-superstar/C77124.html').text
bs = BeautifulSoup(html)
possible_links = bs.find_all('img')
for link in possible_links:
if link.has_attr('src'):
if link.has_key('src'):
if 'zoom' in link['src']:
print link['src']
but im search a metod to scrape a list in automatic
thankyou so much
i try to do this for have list :
from bs4 import BeautifulSoup
import requests
url = "http://www.adidas.it/on/demandware.static/-/Sites-adidas-IT-Library/it_IT/v/sitemap/product/adidas-IT-it-it-product.xml"
r = requests.get(url)
data = r.text
soup = BeautifulSoup(data)
for url in soup.findAll("loc"):
print url.text
but i cant to attach request..
i can find the word "Zoom" in any link present in sitemap.xml
thankyou so much

import requests
from bs4 import BeautifulSoup
import re
def make_soup(url):
r = requests.get(url)
soup = BeautifulSoup(r.text, 'lxml')
return soup
# put urls in a list
def get_xml_urls(soup):
urls = [loc.string for loc in soup.find_all('loc')]
return urls
# get the img urls
def get_src_contain_str(soup, string):
srcs = [img['src']for img in soup.find_all('img', src=re.compile(string))]
return srcs
if __name__ == '__main__':
xml = 'http://www.adidas.it/on/demandware.static/-/Sites-adidas-IT-Library/it_IT/v/sitemap/product/adidas-IT-it-it-product.xml'
soup = make_soup(xml)
urls = get_xml_urls(soup)
# loop through the urls
for url in urls:
url_soup = make_soup(url)
srcs = get_src_contain_str(url_soup, 'zoom')
print(srcs)

Web crawler - following links

Please bear with me. I am quite new at Python - but having a lot of fun. I am trying to code a web crawler that crawls through election results from the last referendum in Denmark. I have managed to extract all the relevant links from the main page. And now I want Python to follow each of the 92 links and gather 9 pieces of information from each of those pages. But I am so stuck. Hope you can give me a hint.
Here is my code:
import requests
import urllib2
from bs4 import BeautifulSoup
# This is the original url http://www.kmdvalg.dk/
soup = BeautifulSoup(urllib2.urlopen('http://www.kmdvalg.dk/').read())
my_list = []
all_links = soup.find_all("a")
for link in all_links:
link2 = link["href"]
my_list.append(link2)
for i in my_list[1:93]:
print i
# The output shows all the links that I would like to follow and gather information from. How do I do that?

Here is my solution using lxml. It's similar to BeautifulSoup
import lxml
from lxml import html
import requests
page = requests.get('http://www.kmdvalg.dk/main')
tree = html.fromstring(page.content)
my_list = tree.xpath('//div[#class="LetterGroup"]//a/#href') # grab all link
print 'Length of all links = ', len(my_list)
my_list is a list consist of all links. And now you can use for loop to scrape information inside each page.
We can for loop through each links. Inside each page, you can extract information as example. This is only for the top table.
table_information = []
for t in my_list:
page_detail = requests.get(t)
tree = html.fromstring(page_detail.content)
table_key = tree.xpath('//td[#class="statusHeader"]/text()')
table_value = tree.xpath('//td[#class="statusText"]/text()') + tree.xpath('//td[#class="statusText"]/a/text()')
table_information.append(zip([t]*len(table_key), table_key, table_value))
For table below the page,
table_information_below = []
for t in my_list:
page_detail = requests.get(t)
tree = html.fromstring(page_detail.content)
l1 = tree.xpath('//tr[#class="tableRowPrimary"]/td[#class="StemmerNu"]/text()')
l2 = tree.xpath('//tr[#class="tableRowSecondary"]/td[#class="StemmerNu"]/text()')
table_information_below.append([t]+l1+l2)
Hope this help!

A simple approach would be to iterate through your list of urls and parse them each individually:
for url in my_list:
soup = BeautifulSoup(urllib2.urlopen(url).read())
# then parse each page individually here
Alternatively, you could speed things up significantly using Futures.
from requests_futures.sessions import FuturesSession
def my_parse_function(html):
"""Use this function to parse each page"""
soup = BeautifulSoup(html)
all_paragraphs = soup.find_all('p')
return all_paragraphs
session = FuturesSession(max_workers=5)
futures = [session.get(url) for url in my_list]
page_results = [my_parse_function(future.result()) for future in results]

This would be my solution for your problem
import requests
from bs4 import BeautifulSoup
def spider():
url = "http://www.kmdvalg.dk/main"
source_code = requests.get(url)
plain_text = source_code.text
soup = BeautifulSoup(plain_text, 'html.parser')
for link in soup.findAll('div', {'class': 'LetterGroup'}):
anc = link.find('a')
href = anc.get('href')
print(anc.getText())
print(href)
# spider2(href) call a second function from here that is similar to this one(making url = to herf)
spider2(href)
print("\n")
def spider2(linktofollow):
url = linktofollow
source_code = requests.get(url)
plain_text = source_code.text
soup = BeautifulSoup(plain_text, 'html.parser')
for link in soup.findAll('tr', {'class': 'tableRowPrimary'}):
anc = link.find('td')
print(anc.getText())
print("\n")
spider()
its not done... i only get a simple element from the table but you get the idea and how its supposed to work.

Here is my final code that works smooth. Please let me know if I could have done it smarter!
import urllib2
from bs4 import BeautifulSoup
import codecs
f = codecs.open("eu2015valg.txt", "w", encoding="iso-8859-1")
soup = BeautifulSoup(urllib2.urlopen('http://www.kmdvalg.dk/').read())
liste = []
alle_links = soup.find_all("a")
for link in alle_links:
link2 = link["href"]
liste.append(link2)
for url in liste[1:93]:
soup = BeautifulSoup(urllib2.urlopen(url).read().decode('iso-8859-1'))
tds = soup.findAll('td')
stemmernu = soup.findAll('td', class_='StemmerNu')
print >> f, tds[5].string,";",tds[12].string,";",tds[14].string,";",tds[16].string,";", stemmernu[0].string,";",stemmernu[1].string,";",stemmernu[2].string,";",stemmernu[3].string,";",stemmernu[6].string,";",stemmernu[8].string,";",'\r\n'
f.close()