Develop Reference

How do i get a list of search response URLS

from bs4 import BeautifulSoup
import requests
searchresults = []
search = 'seo'
url = 'https://www.google.com/search'
headers = {
'Accept' : '*/*',
'Accept-Language': 'en-US,en;q=0.5',
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/98.0.4758.82',
}
parameters = {'q': search}
content = requests.get(url, headers = headers, params = parameters).text
soup = BeautifulSoup(content, 'html.parser')
search = soup.find(id = 'search')
first_link = search.find('a')
searchresults.append(first_link['href'])
for i,j in enumerate(searchresults):
print(searchresults[i])
How do i return the whole search result URL list? I would like to later on add multiple pages soo i can index all the URLs

If you want to get all the links from the search result, replace your code after search = soup.find(id = 'search'):
a_tags = search.find_all('a', href=True)
searchresults = [i['href'] for i in a_tags]
for i,j in enumerate(searchresults):
print(j)
Your code currently gives one 1 link because you are using search.find('a') which gives the first result, instead of search.find_all('a', href=True), which gives all the a tags that have a link.

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
[...]

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

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']

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/

Intermittent BeautifulSoup with Amazon Book ISBN

I am trying to gather some information about some books available on Amazon and I am having a weird glitch error that I can't understand. At first I thought it was Amazon blocking my connection but then I noticed the request has a "200 OK" and it had the real HTML content of the corresponding page.
Let's take for example this book: https://www.amazon.co.uk/All-Rage-Cara-Hunter/dp/0241985110
headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.3'}
url = 'https://www.amazon.co.uk/All-Rage-Cara-Hunter/dp/0241985110/ref=sr_1_1?crid=2PPCQEJD706VY&dchild=1&keywords=books+bestsellers+2020+paperback&qid=1598132071&sprefix=book%2Caps%2C234&sr=8-1'
response = requests.get(url, headers=headers)
soup = BeautifulSoup(response.content, features="lxml")
price = {}
if soup.select("#buyBoxInner > ul > li > span > .a-text-strike") != []:
price["regular_price"] = float(
soup.select("#buyBoxInner > ul > li > span > .a-text-strike")[0].string[1:].replace(",", "."))
price["promo_price"] = float(soup.select(".offer-price")[0].string[1:].replace(",", "."))
else:
price["regular_price"] = float(soup.select(".offer-price")[0].string[1:].replace(",", "."))
price["currency"] = soup.select(".offer-price")[0].string[0]
This part works fine and I can have the regular price and a promo price (if exists), and even the currency. But when I do this:
isbn = soup.select("td.bucket > .content > ul > li")[4].contents[1].string.strip().replace("-", "")
I get "IndexError: list index out of range". But if I debug the code, the content is actually there!
Is this a bug of BeautifulSoup? Is the request response too long?

It seems that Amazon returns two version of the page. One where's <td class="bucket"> and one where are several <span> tags. This script tries to extract ISBN from both of them:
import requests
from bs4 import BeautifulSoup
headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.3'}
url = 'https://www.amazon.co.uk/All-Rage-Cara-Hunter/dp/0241985110'
response = requests.get(url, headers=headers)
soup = BeautifulSoup(response.content, features="lxml")
isbn_10 = soup.select_one('span.a-text-bold:contains("ISBN-10"), b:contains("ISBN-10")').find_parent().text
isbn_13 = soup.select_one('span.a-text-bold:contains("ISBN-13"), b:contains("ISBN-13")').find_parent().text
print(isbn_10.split(':')[-1].strip())
print(isbn_13.split(':')[-1].strip())
Prints:
0241985110
978-0241985113

I wish I had an explanation of the problem but you a solution would be to wrap your code in a function like so:
def scrape():
headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.3'}
url = 'https://www.amazon.co.uk/All-Rage-Cara-Hunter/dp/0241985110/ref=sr_1_1?crid=2PPCQEJD706VY&dchild=1&keywords=books+bestsellers+2020+paperback&qid=1598132071&sprefix=book%2Caps%2C234&sr=8-1'
response = requests.get(url, headers=headers)
soup = BeautifulSoup(response.content, features="lxml")
price = {}
if soup.select("#buyBoxInner > ul > li > span > .a-text-strike") != []:
price["regular_price"] = float(
soup.select("#buyBoxInner > ul > li > span > .a-text-strike")[0].string[1:].replace(",", "."))
price["promo_price"] = float(soup.select(".offer-price")[0].string[1:].replace(",", "."))
else:
price["regular_price"] = float(soup.select(".offer-price")[0].string[1:].replace(",", "."))
price["currency"] = soup.select(".offer-price")[0].string[0]
#ADD THIS FEATURE TO YOUR CODE
isbn = soup.select("td.bucket > .content > ul > li")
if not isbn:
scrape()
isbn = isbn[4].contents[1].string.strip().replace("-", "")
Then if it fails it will just call itself again. You might want to refactor it so it only makes the request once.