I have a script that loads a page and saves a bunch of data ids from multiple containers. I then want to open up new urls appending those said data ids onto the end of the urls. For each url I want to locate all the hrefs and compare them to a list of specific links and if any of them match I want to save that link and a few other details to a table.
I have managed to get it to open the url with the appended data id but when I try to search for elements in the new page it either pulls them from the first url that was parsed if I try to findAll from soup again or I constantly get this error when I try to run another html.parser.
ResultSet object has no attribute 'findAll'. You're probably treating a list of items like a single item. Did you call find_all() when you meant to call find()?
Is it not possible to run another parser or am I just doing something wrong?
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from bs4 import BeautifulSoup as soup
from selenium.webdriver.common.action_chains import ActionChains
url = "http://csgo.exchange/id/76561197999004010#x"
driver = webdriver.Firefox()
driver.get(url)
import time
time.sleep(15)
html = driver.page_source
soup = soup(html, "html.parser")
containers = soup.findAll("div",{"class":"vItem"})
print(len(containers))
data_ids = [] # Make a list to hold the data-id's
for container in containers:
test = container.attrs["data-id"]
data_ids.append(test) # add data-id's to the list
print(str(test))
for id in data_ids:
url2 = "http://csgo.exchange/item/" + id
driver.get(url2)
import time
time.sleep(2)
soup2 = soup(html, "html.parser")
containers2 = soup2.findAll("div",{"class":"bar"})
print(str(containers2))
with open('scraped.txt', 'w', encoding="utf-8") as file:
for id in data_ids:
file.write(str(id)+'\n') # write every data-id to a new line