How to use lists outside of for loop? - python

I´m trying to monitor prices on the eshop. How I do it is, I´m webscraping the prices, then wait some time, do it again and compare the lists with prices if anything changed. But my lists with prices are inside the for loops and when I want to compare them outside the for loop, it says my lists are undefined. I hope you´ll understand that better with my code.
import requests, smtplib, time
from bs4 import BeautifulSoup
url = "https://www.sportisimo.sk/panska-vysoka-trekova-obuv/?riadenie=najlacnejs%C3%AD&dostupnost=vsetko&znacka[]=28"
getPage = requests.get(url)
page_soup = BeautifulSoup(getPage.content, "html.parser")
old_containers = page_soup.findAll("div", {"class": "product-box__in"})
for old_container in old_containers:
price_container = old_container.findAll("span", {"class":"price"})
product_price_old = price_container[0].text.strip()
print(product_price_old)
print("---------------------")
time.sleep(10)
getPage_new = requests.get(url)
page_soup_new = BeautifulSoup(getPage_new.content, "html.parser")
new_containers = page_soup_new.findAll("div", {"class": "product-box__in"})
for new_container in new_containers:
price_container = new_container.findAll("span", {"class": "price"})
product_price_new = price_container[0].text.strip()
print(product_price_new)
if product_price_old != product_price_new:
for new_container in new_containers:
name_container = new_container.findAll("div", {"class": "product-box__name"})
product_name = name_container[0].h2.a.text.strip()
print(f"Zmenený produkt: {product_name}")
print(f"Pôvodná cena: {product_price_old}")
print(f"Nová cena: {product_price_new}")
The product_price_old and product_price_new writes me that they are undefined lists. How can I fix this to work, please?
Thank you

You could define start with
product_price_old = []
product_price new = []
and in the loops you could do
for element in price_container[0].text.strip():
product_price_old.append(element)
# or whatever format you want your data in

Related

Function to get multiple soups only gives one soup

so I am trying to create a function that creates a list of multiple soups. I started with doing it in normal code (I do not know how to call this exactly:
list_url = ["http://www.facebook.com", "https://www.google.com", "http://www.yahoo.com"]
list_soup = []
for url in list_url:
soup = BeautifulSoup(requests.get(url).text, "html.parser")
list_soup.append(soup)
And this code works, but when I a function of this:
def get_multi_soup(list_url):
list_multi = []
for url in list_url:
soup = BeautifulSoup(requests.get(url).text, "html.parser")
list_multi.append(soup)
return list_multi
list_soup = get_multi_soup(list_url)
The code does not work as intended as it only gives one soup instead of three.
Can somebody explain why this does not work? The list_soup equals only one soup.
just move the return outside the for loop.
You are returning on the first iteration.
def get_multi_soup(list_url):
list_multi = []
for url in list_url:
soup = BeautifulSoup(requests.get(url).text, "html.parser")
list_multi.append(soup)
return list_multi
list_soup = get_multi_soup(list_url)
Should do the trick. :)

web-scrape: get H4 attributes & href

I am trying to web-scrape a website. But I can get access to the attributes of some fields.
here is the code i used:
import urllib3
from bs4 import BeautifulSoup
import pandas as pd
scrap_list = pd.DataFrame()
for path in range(10): # scroll over the categories
for path in range(10): # scroll over the pages
url = 'https://www.samehgroup.com/index.php?route=product/category'+str(page)+'&'+'path='+ str(path)
req = urllib3.PoolManager()
res = req.request('GET', URL)
soup = BeautifulSoup(res.data, 'html.parser')
soup.findAll('h4', {'class': 'caption'})
# extract names
scrap_name = [i.text.strip() for i in soup.findAll('h2', {'class': 'caption'})]
scrap_list['product_name']=pd.DataFrame(scrap_name,columns =['Item_name'])
# extract prices
scrap_list['product_price'] = [i.text.strip() for i in soup.findAll('div', {'class': 'price'})]
product_price=pd.DataFrame(scrap_price,columns =['Item_price'])
I want an output that provides me with each product and its price. I still can't get that right.
Any help would be very much appreciated.
I think the problem here was looping through the website pages. I got the code below working by first making a list of urls containing numbered 'paths' corresponding to pages on the website. Then looping through this list and applying a page number to the url.
If you wanted to only get all the products from a certain page, this page can be selected from the urlist and by index.
from bs4 import BeautifulSoup
import requests
import pandas as pd
import time
urlist = [] #create list of usable url's to iterate through,
for i in range(1,10): # 9 pages equal to pages on website
urlist.append('https://www.samehgroup.com/index.php?route=product/category&path=' + str(i))
namelist = []
newprice = []
for urlunf in urlist: #first loop to get 'path'
for n in range(100): #second loop to get 'pages'. set at 100 to cover website max page at 93
try: #try catches when pages containing products run out.
url = urlunf + '&page=' + str(n)
page = requests.get(url).text
soup = BeautifulSoup(page, 'html')
products = soup.find_all('div', class_='caption')
for prod in products: #loops over returned list of products for names and prices
name = prod.find('h4').text
newp = prod.find('p', class_='price').find('span', class_='price-new').text
namelist.append(name) #append data to list outside of loop
newprice.append(newp)
time.sleep(2)
except AttributeError: #if there are no more products it will move to next page
pass
df = pd.DataFrame() #create df and add scraped data
df['name'] = namelist
df['price'] = newprice

Use Beatiful Soup in scraping multiple websites

I want to know why lists all_links and all_titles don't want to receive any records from lists titles and links. I have tried also .extend() method and it didn't help.
import requests
from bs4 import BeautifulSoup
all_links = []
all_titles = []
def title_link(page_num):
page = requests.get(
'https://www.gumtree.pl/s-mieszkania-i-domy-sprzedam-i-kupie/warszawa/page-%d/v%dc9073l3200008p%d'
% (page_num, page_num, page_num))
soup = BeautifulSoup(page.content, 'html.parser')
links = ['https://www.gumtree.pl' + link.get('href')
for link in soup.find_all('a', class_ ="href-link tile-title-text")]
titles = [flat.next_element for flat in soup.find_all('a', class_ = "href-link tile-title-text")]
print(titles)
for i in range(1,5+1):
title_link(i)
all_links = all_links + links
all_titles = all_titles + titles
i+=1
print(all_links)
import pandas as pd
df = pd.DataFrame(data = {'title': all_titles ,'link': all_links})
df.head(100)
#df.to_csv("./gumtree_page_1.csv", sep=';',index=False, encoding = 'utf-8')
#df.to_excel('./gumtree_page_1.xlsx')
When I ran your code, I got
NameError Traceback (most recent call last)
<ipython-input-3-6fff0b33d73b> in <module>
16 for i in range(1,5+1):
17 title_link(i)
---> 18 all_links = all_links + links
19 all_titles = all_titles + titles
20 i+=1
NameError: name 'links' is not defined
That points to a problem - variable named links is not defined in a global scope (where you add it to all_links). You can read about python scopes here. You'd need to return links and titles from title_link. Something similar to this:
def title_link(page_sum):
# your code here
return links, titles
for i in range(1,5+1):
links, titles = title_link(i)
all_links = all_links + links
all_titles = all_titles + titles
print(all_links)
This code is exhibits confusion about scoping. titles and links inside of title_link are local to that function. When the function ends, the data disappears and it cannot be accessed from another scope such as main. Use the return keyword to return values from functions. In this case, you'd need to return a tuple pair of titles and links like return titles, links.
Since functions should do one task only, having to return a pair shows reveals a possible design flaw. A function like title_link is overloaded and should probably be two separate functions, one to get titles and one to get links.
Having said that, the functions here seem like premature abstractions since the operations can be done directly.
Here's a suggested rewrite:
import pandas as pd
import requests
from bs4 import BeautifulSoup
url = "https://www.gumtree.pl/s-mieszkania-i-domy-sprzedam-i-kupie/warszawa/page-%d/v%dc9073l3200008p%d"
data = {"title": [], "link": []}
for i in range(1, 6):
page = requests.get(url % (i, i, i))
soup = BeautifulSoup(page.content, "html.parser")
titles = soup.find_all("a", class_="href-link tile-title-text")
data["title"].extend([x.next_element for x in titles])
data["link"].extend("https://www.gumtree.pl" + x.get("href") for x in titles)
df = pd.DataFrame(data)
print(df.head(100))
Other remarks:
i+=1 is unnecessary; for loops move forward automatically in Python.
(1,5+1) is clearer as (1, 6).
List comprehensions are great, but if they run multiple lines, consider writing them as normal loops or creating an intermediate variable or two.
Imports should be at the top of a file only. See PEP-8.
list.extend(other_list) is preferable to list = list + other_list, which is slow and memory-intensive, creating a whole copy of the list.
Try this:
import requests
from bs4 import BeautifulSoup
all_links = []
all_titles = []
def title_link(page_num):
page = requests.get(
'https://www.gumtree.pl/s-mieszkania-i-domy-sprzedam-i-kupie/warszawa/page-%d/v%dc9073l3200008p%d'
% (page_num, page_num, page_num))
page.encoding = 'utf-8'
soup = BeautifulSoup(page.content, 'html.parser', from_encoding='utf-8')
links = ['https://www.gumtree.pl' + link.get('href')
for link in soup.find_all('a', class_ ="href-link tile-title-text")]
titles = [flat.next_element for flat in soup.find_all('a', class_ = "href-link tile-title-text")]
print(titles)
return links, titles
for i in range(1,5+1):
links, titles = title_link(i)
all_links.extend(links)
all_titles.extend(titles)
# i+=1 not needed in python
print(all_links)
import pandas as pd
df = pd.DataFrame(data = {'title': all_titles ,'link': all_links})
df.head(100)
I think you just needed to get links and titles out of title_link(page_num).
Edit: removed the manual incrementing per comments
Edit: changed the all_links = all_links + links to all_links.extend(links)
Edit: website is utf-8 encoded, added page.encoding = 'utf-8' and as extra (probably unnecessary) measure, from_encoding='utf-8' to the BeautifulSoup

Web Scraping with Python - blank return

I'm trying to scrape reviews from TrustPilot, but the code always return with blank sheets and the headers/categories I specified. Could someone help me with this?
from bs4 import BeautifulSoup, SoupStrainer
import pandas as pd
driver= webdriver.Chrome()
names=[] #List to store name of the product
headers=[] #List to store price of the product
bodies=[]
ratings=[] #List to store rating of the product
dates=[]
#driver = webdriver.Chrome(ChromeDriverManager().install())
driver.get("https://www.trustpilot.com/review/birchbox.com?page=2")
content = driver.page_source
soup = BeautifulSoup(content, "html.parser", parse_only=SoupStrainer('a'))
for a in soup.findAll('a', href=True, attrs={'class':'reviews-container'}):
name=a.find('div', attrs={'class':'consumer-information_name'})
header=a.find('div', attrs={'class':'review-content_title'})
body=a.find('div', attrs={'class':'review-content_text'})
rating=a.find('div', attrs={'class':'star-rating star-rating--medium'})
date=a.find('div', attrs={'class':'review-date--tooltip-target'})
names.append(name.text)
headers.append(header.text)
bodies.append(body.text)
ratings.append(rating.text)
dates.append(date.text)
print ('webpage, no errors')
df = pd.DataFrame({'User Name':names,'Header':headers,'Body':bodies,'Rating':ratings,'Date':dates})
df.to_csv('reviews02.csv', index=False, encoding='utf-8')
print ('csv made')```
The issue is soup.findAll('a', href=True, attrs={'class':'reviews-container'}) is not finding any results, so there are 0 iterations in the loop. Make sure you are using the correct tags and class names. Also you don't need to use a loop because BeautifulSoup has a find_all method. I used the requests module to open the web page, though it shouldn't make a difference.
from bs4 import BeautifulSoup
import requests
req = requests.get("https://www.trustpilot.com/review/birchbox.com?page=2")
content = req.content
soup = BeautifulSoup(content, "lxml")
names = soup.find_all('div', attrs={'class': 'consumer-information__name'})
headers = soup.find_all('h2', attrs={'class':'review-content__title'})
bodies = soup.find_all('p', attrs={'class':'review-content__text'})
ratings = soup.find_all('div', attrs={'class':'star-rating star-rating--medium'})
dates = soup.find_all('div', attrs={'class':'review-content-header__dates'})
And now each list has 20 entries.

Parsing multiple urls with Python and BeautifulSoup

I started learning Python today and so it is not a surprise that I am struggling with some basics. I am trying to parse data from a school website for a project and I managed to parse the first page. However, there are multiple pages (results are paginated).
I have an idea about how to go about it, ie, run through the urls in a loop since I know the url format but I have no idea how to proceed. I figured it would be better to somehow search for the "next" button and run the function if it is there, if not, then stop function.
I would appreciate any help I can get.
import requests
from bs4 import BeautifulSoup
url = "http://www.myschoolwebsite.com/1"
#url2 = "http://www.myschoolwebsite.com/2"
r = requests.get(url)
soup = BeautifulSoup(r.content,'lxml')
g_data = soup.find_all('ul', {"class": "searchResults"})
for item in g_data:
for li in item.findAll('li'):
for resultnameh2 in li.findAll('h2'):
for resultname in resultnameh2.findAll('a'):
print(resultname).text
for resultAddress in li.findAll('p', {"class": "resultAddress"}):
print(resultAddress).text.replace('Get directions','').strip()
for resultContact in li.findAll('ul', {"class": "resultContact"}):
for resultContact in li.findAll('a', {"class": "resultMainNumber"}):
print(resultContact).text
First, you can assume the maximum no. of pages of the directory (if you know pattern of the url). I am assuming the url is of the form http://base_url/page Next you can write this:
base_url = 'http://www.myschoolwebsite.com'
total_pages = 100
def parse_content(r):
soup = BeautifulSoup(r.content,'lxml')
g_data = soup.find_all('ul', {"class": "searchResults"})
for item in g_data:
for li in item.findAll('li'):
for resultnameh2 in li.findAll('h2'):
for resultname in resultnameh2.findAll('a'):
print(resultname).text
for resultAddress in li.findAll('p', {"class": "resultAddress"}):
print(resultAddress).text.replace('Get directions','').strip()
for resultContact in li.findAll('ul', {"class": "resultContact"}):
for resultContact in li.findAll('a', {"class": "resultMainNumber"}):
print(resultContact).text
for page in range(1, total_pages):
response = requests.get(base_url + '/' + str(page))
if response.status_code != 200:
break
parse_content(response)
I would make an array with all the URLs and loop through it, or if there is a clear pattern, write a regex to search for that pattern.

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