How can I get the data from a span in BeautifulSoup? - python

This is my code, I want to take the location's name and link, the variable "lugares" finds multiple item-containers, but I only want the first one [0]; then goes the for loop, but I can't find the span classes.
from bs4 import BeautifulSoup
import requests
b=[]
i="https://www.vivanuncios.com.mx"
url = "https://www.vivanuncios.com.mx/s-renta-inmuebles/estado-de-mexico/v1c1098l1014p1"
encabezado = {"User-Agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/94.0.4606.71 Safari/537.36",'Accept-Language': 'en-US, en;q=0.5'}
page =requests.get(url,headers=encabezado)
soup = BeautifulSoup(page.content,"html.parser")
lugares = soup.find_all("div",{"class":"items-container"})
lugares=lugares[0]
print(len(lugares))
for lugar in lugares:
locationlink = i + str(lugar.find("span",{"class":"item"}).find("a")["href"])
location= lugar.find("span",{"class":"item"}).text
a=[location,locationlink]
b.append(a)

There are multiple options to get the goal, best one depence on what you expect and wanna do with this information in follow up process.
First Option
If you are just looking for the infos of first location you can do the following:
lugar = soup.select_one('div.items-container a')
b = [lugar.text, f'{i}{lugar["href"]}']
or
lugar = soup.select('div.items-container a')[0]
b = [lugar.text, f'{i}{lugar["href"]}']
Both select the first <a> in the <div> with class items-container.
Output
['Huixquilucan','https://www.vivanuncios.com.mx/s-renta-inmuebles/huixquilucan/v1c1098l10689p1']
Alternativ
If you are interested to get all at once, you should use a list of dicts, so later on you just have to iterate it and get all information in place:
[{'name':x.text, 'link':f'{i}{x["href"]}'} for x in soup.select('div.items-container a')]
Output
[{'name': 'Huixquilucan',
'link': 'https://www.vivanuncios.com.mx/s-renta-inmuebles/huixquilucan/v1c1098l10689p1'},
{'name': 'Naucalpan',
'link': 'https://www.vivanuncios.com.mx/s-renta-inmuebles/naucalpan/v1c1098l10710p1'},
{'name': 'Atizapán',
'link': 'https://www.vivanuncios.com.mx/s-renta-inmuebles/atizapan/v1c1098l10662p1'},
{'name': 'Metepec',
'link': 'https://www.vivanuncios.com.mx/s-renta-inmuebles/metepec-edomex/v1c1098l10707p1'},...]
Example (showing results of both)
from bs4 import BeautifulSoup
import requests
i="https://www.vivanuncios.com.mx"
url = "https://www.vivanuncios.com.mx/s-renta-inmuebles/estado-de-mexico/v1c1098l1014p1"
encabezado = {"User-Agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/94.0.4606.71 Safari/537.36",'Accept-Language': 'en-US, en;q=0.5'}
page =requests.get(url,headers=encabezado)
soup = BeautifulSoup(page.content,"html.parser")
lugar = soup.select_one('div.items-container a')
b = [lugar.text, f'{i}{lugar["href"]}']
print(f'First lugar:\n {b} \n')
## or alternative option
allLugaros = [{'name':x.text, 'link':f'{i}{x["href"]}'} for x in soup.select('div.items-container a')]
print(f'First lugar from lugaros (list of dict):\n {allLugaros[0]} \n')
print(f'All lugaros as list of dict:\n {allLugaros} \n')

First, you need to get all spans in the first Lugares lugares[0].
Then you need to iterate for each span to get the link and text for each location.
The Code:
from bs4 import BeautifulSoup
import requests
b=[]
i="https://www.vivanuncios.com.mx"
url = "https://www.vivanuncios.com.mx/s-renta-inmuebles/estado-de-mexico/v1c1098l1014p1"
encabezado = {"User-Agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/94.0.4606.71 Safari/537.36",'Accept-Language': 'en-US, en;q=0.5'}
page =requests.get(url,headers=encabezado)
soup = BeautifulSoup(page.content,"html.parser")
lugares = soup.find_all("div",{"class":"items-container"})
#lugares=lugares[0]
print(len(lugares))
# get all spans
spans = lugares[0].find_all("span",{"class":"item"})
# itreate throw each span
for span in spans:
# get location text
location = span.find("a").text
# locationlink builder
site = "www.vivanuncios.com.mx"
link = span.find("a")["href"]
locationlink = f"{site}{link}"
a = [location,locationlink]
b.append(a)
print (b[0])
Output:
['Huixquilucan', 'www.vivanuncios.com.mx/s-renta-inmuebles/huixquilucan/v1c1098l10689p1']

Related

I am trying to navigate through the pages of a website and scrape its links but the same page data is scraped even after changing page number

from bs4 import BeautifulSoup
import requests
import pymongo
def traverse_source():
article_links = []
for pgindx in range(9):
headers = {
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.36',
"path": f"issue/S0196-0644(21)X0012-1?pageStart={pgindx}",
"Sec-fetch-site": "same-origin",
}
source_url = ""
source_data = requests.get(source_url,headers = headers)
print(source_data.headers)
source_url = None
source_soup = BeautifulSoup(source_data.content,"html.parser")
destination = source_soup.find_all("h3",attrs = {'class': 'toc__item__title' })
for dest in destination:
try:
article_links.append("https://www.annemergmed.com"+dest.a['href'])
except:
pass
source_soup = None
print(article_links)
if __name__ == "__main__":
traverse_source()
Here even after incrementing the page number in the URL, the content of the first webpage is always scraped. I tried navigating through the pages using GET method (changing the URL) but still even after changing the source url, it is still scraping the data of page number 1
This is one way of scraping that data:
import pandas as pd
import requests
from bs4 import BeautifulSoup
from tqdm import tqdm
headers = {
"User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/103.0.0.0 Safari/537.36"
}
s = requests.Session()
s.headers.update(headers)
big_list = []
for x in tqdm(range(9)):
r = s.get(f'https://www.annemergmed.com/issue/S0196-0644(21)X0012-1?pageStart={x}')
soup = BeautifulSoup(r.text, 'html.parser')
titles = soup.select('div.articleCitation')
for t in titles:
url = t.select_one('h3 a').get('href')
header = t.select_one('h3 a').text
try:
authors = t.select_one('ul.toc__item__authors').get_text(strip=True)
except Exception as e:
authors = 'Unknown'
big_list.append((header, f'https://www.annemergmed.com{url}', authors))
df = pd.DataFrame(list(set(big_list)), columns = ['Title', 'Url', 'Authors'])
print(df.shape)
print(df.head(50))
This will return:
(409, 3)
Title Url Authors
0 194 Challenging the Dogma of Radiographs a Joint Above and Below a Suspected Fracture: Quantification of Waste in Wrist Fracture Evaluation https://www.annemergmed.com/article/S0196-0644(21)01046-5/fulltext M. Rozum,D. Mark Courtney,D. Diercks,S. McDonald
1 112 A Geographical Analysis of Access to Trauma Care From US National Parks in 2018 https://www.annemergmed.com/article/S0196-0644(21)00963-X/fulltext S. Robichaud,K. Boggs,B. Bedell,...A. Sullivan,N. Harris,C. Camargo
2 87 Emergency Radiology Overreads Change Management of Transferred Patients With Traumatic Injuries https://www.annemergmed.com/article/S0196-0644(21)00937-9/fulltext M. Vrablik,R. Kessler,M. Vrablik,...J. Robinson,D. Hippe,M. Hall
[...]

Beautiful soup web scraping returning None-Python

I have a list of movies that I want to scrap the genres from Google.
I've built this code:
import requests
from bs4 import BeautifulSoup
list=['Se7en','Cinema Paradiso','The Shining','Toy Story 3','Capernaum']
gen2 = {}
for i in list:
user_query = i +'movie genre'
URL = 'https://www.google.co.in/search?q=' + user_query
headers = {'User-Agent':'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/102.0.5005.63 Safari/537.36'}
page = requests.get(URL, headers=headers)
soup = BeautifulSoup(page.content, 'html.parser')
c = soup.find(class_='EDblX DAVP1')
print(c)
if c != None:
genres = c.findAll('a')
gen2[i]= genres
But it returns an empty dict, so I checked one by one and it worked, for example:
import requests
from bs4 import BeautifulSoup
user_query = 'Se7en movie genre'
URL = "https://www.google.co.in/search?q=" + user_query
headers = {'User-Agent':'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/102.0.5005.63 Safari/537.36'}
page = requests.get(URL, headers=headers)
soup = BeautifulSoup(page.content, 'html.parser')
v = soup.find(class_='KKHQ8c')
h = {}
genres = v.findAll('a')
for genre in genres:
h['Se7en']=genre
So I find out that in the for loop the variable c is returning None.
I can't figure out why! It only return None inside the loop.
Currently, your URLs are of the form
URLs
so the returned results(google) aren't accurate for all the movies.
You can change it to
`for i in list:
i="+".join(i.split(" "));
user_query = i + "+movie+genre"
URL = 'https://www.google.com/search?q=+'+user_query`
also, movies that belong to a single genre like Cinema Paradiso are in a div with class name "Z0LcW".

How to get all URLs within a page fom oddsportal?

I have a code that scrapes all URLs from oddsportal.com main page.
I want the subsequent links to all pages within the parent URL
e.g.
https://www.oddsportal.com/soccer/africa/africa-cup-of-nations/results/
has further pages i.e. https://www.oddsportal.com/soccer/africa/africa-cup-of-nations/results/, https://www.oddsportal.com/soccer/africa/africa-cup-of-nations-2019/results/, etc.
How can I get that?
My existing code:
import requests
import bs4 as bs
import pandas as pd
url = 'https://www.oddsportal.com/results/#soccer'
headers = {
'User-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/91.0.4472.114 Safari/537.36'}
resp = requests.get(url, headers=headers)
soup = bs.BeautifulSoup(resp.text, 'html.parser')
base_url = 'https://www.oddsportal.com'
a = soup.findAll('a', attrs={'foo': 'f'})
# This set will have all the URLs of the main page
s = set()
for i in a:
s.add(base_url + i['href'])
s = list(s)
# This will filter for all soccer URLs
s = [x for x in s if '/soccer/' in x]
s = pd.DataFrame(s)
print(s)
I am very new to webscraping and hence this question.
You can find main_div tag based on class attribute and use find_all method to get a tag by looping over it you can extract href of it
from bs4 import BeautifulSoup
import requests
headers = {
'User-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/91.0.4472.114 Safari/537.36'}
source = requests.get("https://www.oddsportal.com/soccer/africa/africa-cup-of-nations/results/",headers=headers)
soup = BeautifulSoup(source.text, 'html.parser')
main_div=soup.find("div",class_="main-menu2 main-menu-gray")
a_tag=main_div.find_all("a")
for i in a_tag:
print(i['href'])
Output:
/soccer/africa/africa-cup-of-nations/results/
/soccer/africa/africa-cup-of-nations-2019/results/
/soccer/africa/africa-cup-of-nations-2017/results/
/soccer/africa/africa-cup-of-nations-2015/results/
/soccer/africa/africa-cup-of-nations-2013/results/
/soccer/africa/africa-cup-of-nations-2012/results/
/soccer/africa/africa-cup-of-nations-2010/results/
/soccer/africa/africa-cup-of-nations-2008/results/

How to grab spot price from yahoo finance using BeautifulSoup

I'm trying to grab the spot price of the SPY ETF: https://finance.yahoo.com/quote/SPY/options
I've mostly tried using soup.find_all, using the nested 'div' tags:
from bs4 import BeautifulSoup
import urllib.request
url = 'https://finance.yahoo.com/quote/SPY/options/'
source = urllib.request.urlopen(url).read()
soup = BeautifulSoup(source,'lxml')
for div in soup.find_all('div', class_ = "My(6px) smartphone_Mt(15px)"):
print(div.text)
for div in soup.find_all('div', class_ = "D(ib) Maw(65%) Ov(h)"):
print(div.text)
for div in soup.find_all('div', class_ = "D(ib) Mend(20px)"):
print(div.text)
Nothing is printed. I also tried the following:
print(soup.find('span', attrs = {'data-reactid':"35"}).text)
which results in 'Last Price' being printed. Now obviously I want the last price, rather than the words 'last price', but this is closer.
Nested in that span tag is some html which includes the number I want. I'm guessing the correct answer has to do with the 'react text: 36' stuff within the span tag (can't type it without stackoverflow thinking I'm trying to actually implement the html into this question).
If you just want the price:
import urllib.request
from bs4 import BeautifulSoup, Comment
page = urllib.request.urlopen("https://finance.yahoo.com/quote/SPY?p=SPY")
content = page.read().decode('utf-8')
soup = BeautifulSoup(content, 'html.parser')
comments = soup.findAll(text=lambda text:isinstance(text, Comment))
[comment.extract() for comment in comments]
price = soup.find("span", {"data-reactid": "14", "class" : "Trsdu(0.3s) "}).text
print(price)
Outputs:
271.40
I recommend to you use scrapy, requests modules
import requests
from bs4 import BeautifulSoup
from scrapy.selector import Selector
ajanlar = [
'Mozilla/5.0 (Windows NT 6.1) AppleWebKit/537.36 (KHTML, like Gecko)',
'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_1) AppleWebKit/537.36 (KHTML, like Gecko)',
'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko)',
'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko)',
'Mozilla/5.0 (Windows NT 6.4; WOW64) AppleWebKit/537.36 (KHTML, like Gecko)',
'Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko)']
url = "https://finance.yahoo.com/quote/SPY/options"
headers = {"User-Agent":random.choice(ajanlar)}
response = requests.get(url,headers=headers,proxies=None)
soup = BeautifulSoup(response.text, 'lxml')
xpath1 = "normalize-space(//div[#class='Mt(6px) smartphone_Mt(15px)'])"
xpath2 = "normalize-space(//div[#class='D(ib) Maw(65%) Maw(70%)--tab768 Ov(h)'])"
xpath3 = "normalize-space(//div[#class='D(ib) Mend(20px)'])"
var1 = Selector(text=response.text).xpath(xpath1).extract()[0]
var2 = Selector(text=response.text).xpath(xpath2).extract()[0]
var3 = Selector(text=response.text).xpath(xpath3).extract()[0]
print(var1)
print(var2)
print(var3)
Outputs:
269.97-1.43 (-0.53%)At close: 4:00PM EST269.61 -0.44 (-0.16%)After hours: 6:08PM ESTPeople also watchDIAIWMQQQXLFGLD
269.97-1.43 (-0.53%)At close: 4:00PM EST269.61 -0.44 (-0.16%)After hours: 6:08PM EST
269.97-1.43 (-0.53%)At close: 4:00PM EST
After than, you could apply regex

Scraping seat layout page of book my show using python

I am trying to scrape the bookmyshow website for finding out movie details like at what time tickets are available and how many seats are available. I have got to find how to get the show timings in which seats are available but now i want to get total seats avaialble in that show. My code is :
import requests
from bs4 import BeautifulSoup
import json
base_url = "https://in.bookmyshow.com"
s =requests.session()
headers = {"user-agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/63.0.3239.132 Safari/537.36"}
r = s.get("https://in.bookmyshow.com/vizag/movies", headers = headers)
print(r.status_code)
soup = BeautifulSoup(r.text,"html.parser")
movies_list = soup.find("div",{"class":"__col-now-showing"})
movies = movies_list.findAll("a",{"class":"__movie-name"})
for movie in movies:
print(movie.text)
show = []
containers = movies_list.findAll("div",{"class":"card-container"})
for container in containers:
try:
detail = container.find("div",{"class":"__name overflowEllipses"})
button = container.find("div",{"class":"book-button"})
print(detail.text)
print(button.a["href"])
url_ticket = base_url + button.a["href"]
show.append(url_ticket)
except:
pass
for i in show:
print(i)
for t in show:
res = s.get(t,headers=headers)
bs = BeautifulSoup(res.text,"html.parser")
movie_name = bs.find("div",{"class":"cinema-name-wrapper"})
print(movie_name.text.replace(" ","").replace("\t","").replace("\n",""))
venue_list = bs.find("ul",{"id":"venuelist"})
venue_names = venue_list.findAll("li",{"class":"list"})
try:
for i in venue_names:
vn = i.find("div",{"class":"__name"})
print(vn.text.replace(" ","").replace("\t","").replace("\n",""))
show_times = i.findAll("div",{"data-online":"Y"})
for st in show_times:
print(st.text.replace(" ","").replace("\t","").replace("\n",""))
except:
pass
print("\n")
heads = {
"accept":"*/*",
"accept-encoding":"gzip, deflate, br",
"accept-language":"en-US,en;q=0.9",
"origin":"https://in.bookmyshow.com",
"referer":"https://in.bookmyshow.com/buytickets/chalo-vizag/movie-viza-ET00064364-MT/20180204",
"user-agent":"Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/63.0.3239.132 Safari/537.36"
}
rr = s.post("https://b-eu.simility.com/b?c=bookmyshow&v=1.905&ec=BLOFaZ2HdToCxwcr&cl=0&si=5a76bfce6ae4a00027767ae9&sc=3B0CB9F4-4A27-4588-9FB4-A2A2760569BC&uc=D834EDA4-57E4-4889-A34F-473AC6BBDDBB&e=Seatlayout&cd=.simility.com&r=0&st=1517731803171&s=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",headers =heads) # i got the link while i was inspecting the booking tickets page
f = s.get("https://in.bookmyshow.com/buytickets/chalo-vizag/movie-viza-ET00064364-MT/20180204#!seatlayout") # this is the page gets displayed when we click the show time
ff = f.text
j = json.loads(ff)
print(j)
After i get the source code of this page i can get seats availability easily. But i am unable to get that page. How to do this? Thanks in Advance!
Steps:
1) use selenium to click on the time showing block
driver.find_element_by_xpath('<enter xpath>').click()
find xpath using inspect element and then click on element then copy you will get the option for copy xpath
time.sleep(4) # wait for 4 seconds for the page to appear
2) Get the html source code using
html = driver.page_source
then use beautiful soup to scrap the page
soup = BeautifulSoup(html,'html.parser')
Find all a href tag having class ='_available' and count them and then
find all a href tag having class = '_blocked' and count them
using these data you can find total no of seats and available seats

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