Which element to use in Selenium? - python

I want to find "Moderat" in <p class="text-spread-level">Moderat</p>
I have tried with id, name, xpath and link text.

Would you like to try this?
from bs4 import BeautifulSoup
import requests
sentences = []
res = requests.get(url) # assign your url in variable
soup = BeautifulSoup(res.text, "lxml")
tag_list = soup.select("p.text-spread-level")
for tag in tag_list:
sentences.append(tag.text)
print(sentences)

Find the element by class name and get the text.
el=driver.find_element_by_class_name('text-spread-level')
val=el.text
print(val)

Related

Get html text with Beautiful Soup

I'm trying to get the number from inside a div:
<div class="tv-symbol-price-quote__value js-symbol-last">122.7<span class="">8</span></div>
I need the 122.7 number, but I cant get it. I have tried with:
strings = soup.find("div", class_="tv-symbol-price-quote__value js-symbol-last").string
But, there are more than one element and I receive "none".
Is there a way to print the childs and get the string from childs?
Use .getText().
For example:
from bs4 import BeautifulSoup
sample_html = """
<div class="tv-symbol-price-quote__value js-symbol-last">122.7<span class="">8</span></div>
"""
soup = BeautifulSoup(sample_html, "html.parser")
strings = soup.find("div", class_="tv-symbol-price-quote__value js-symbol-last").getText()
print(strings)
Output:
122.78
Or use __next__() to get only the 122.7.
soup = BeautifulSoup(sample_html, "html.parser")
strings = soup.find("div", class_="tv-symbol-price-quote__value js-symbol-last").strings.__next__()
print(strings)
Output:
122.7
To only get the first text, search for the tag, and call the next_element method.
from bs4 import BeautifulSoup
html = """
<div class="tv-symbol-price-quote__value js-symbol-last">122.7<span class="">8</span></div>
"""
soup = BeautifulSoup(html, "html.parser")
print(
soup.find("div", class_="tv-symbol-price-quote__value js-symbol-last").next_element
)
Output:
122.7
You could use selenium to find the element and then use BS4 to parse it.
An example would be
import selenium.webdriver as WD
from selenium.webdrive.chrome.options import Options
import bs4 as B
driver = WD.Chrome()
objXpath = driver.find_element_by_xpath("""yourelementxpath""")
objHtml = objXpath.get_attribute("outerHTML")
soup = B.BeutifulSoup(objHtml, 'html.parser')
text = soup.get_text()
This code should work.
DISCLAIMER
I haven't done work w/ selenium and bs4 in a while so you might have to tweak it a little bit.

Grabbing first link in this code with Python

Hello this is the code I want to grab first link from using BeautifulSoup.
view-source:https://www.binance.com/en/blog
I want to grab the first article here so it would be "Trust Wallet Now Supports Stellar Lumens, 4 More Tokens"
I am trying to use Python for this.
I use this code but it grabs all the links, I only want first one to grab
with open('binanceblog1.html', 'w') as article:
before13 = requests.get("https://www.binance.com/en/blog", headers=headers2)
data1b = before13.text
xsoup2 = BeautifulSoup(data1b, "lxml")
for div in xsoup2.findAll('div', attrs={'class':'title sc-0 iaymVT'}):
before_set13 = div.find('a')['href']
How can I do this?
Most simple solution I can think at this moment that works with your code is to use break, this is because of findAll
for div in xsoup2.findAll('div', attrs={'class':'title sc-62mpio-0 iIymVT'}):
before_set13 = div.find('a')['href']
break
For just the first element you can use find
before_set13 = soup.find('div', attrs={'class':'title sc-62mpio-0 iIymVT'}).find('a')['href']
Try (Extracting the href from 'Read more' button)
import requests
from bs4 import BeautifulSoup
r = requests.get('https://www.binance.com/en/blog')
soup = BeautifulSoup(r.text, "html.parser")
div = soup.find('div', attrs={'class': 'read-btn sc-62mpio-0 iIymVT'})
print(div.find('a')['href'])
You can assess the situation inside the loop and break when you find a satisfactory result.
for div in xsoup2.findAll('div', attrs={'class':'title sc-62mpio-0 iIymVT'}):
before_set13 = div.find('a')['href']
if before_set13 != '/en/blog':
break
print('skipping ' + before_set13)
print('grab ' + before_set13)
Output of the code with these changes:
skipping /en/blog
grab /en/blog/317619349105270784/Trust-Wallet-Now-Supports-Stellar-Lumens-4-More-Tokens
Use the class name with class css selector (.) for the content section then descendant combinator with a type css selector to specify child a tag element. select_one returns first match
soup.select_one('.content a')['href']
Code:
from bs4 import BeautifulSoup as bs
import requests
r = requests.get('https://www.binance.com/en/blog')
soup = bs(r.content, 'lxml')
link = soup.select_one('.content a')['href']
print('https://www.binance.com' + link)

Access attributes with beautifulSoup and print

I'd like to scrape a site to findall title attributes of h2 tag
<h2 class="1">Titanic_Caprio</h2>
Using this code, I'm accessing the entire h2 tag
from bs4 import BeautifulSoup
import urllib2
url = "http://www.example.it"
page = urllib2.urlopen(url)
soup = BeautifulSoup(page, 'html.parser')
links = soup.findAll('h2')
print "".join([str(x) for x in links] )
using findAll('h2', attrs = {'title'}) doesn't have results. What Am I doing wrong? How can I print out the entire title's list in a file?
The problem is that title is not an attribute of the h2 tag, but of a tag included in it. So you must first search for <h2> tags, and then subtags having a title attribute:
titles = []
h2_list = links = soup.findAll('h2')
for h2 in h2_list:
titles.extend(h2.findAll(lambda x: x.has_attr('title')))
It works because BeautifulSoup can use functions as search filters.
you need to pass key value pairs in attrs
findAll('h2', attrs = {"key":"value"})

How to extract href links from anchor tags using BeautifulSoup?

I've been trying to extract just the links corresponding to the jobs on each page. But for some reason they dont print when I execute the script. No errors occur.
for the inputs I put engineering, toronto respectively. Here is my code.
import requests
from bs4 import BeautifulSoup
import webbrowser
jobsearch = input("What type of job?: ")
location = input("What is your location: ")
url = ("https://ca.indeed.com/jobs?q=" + jobsearch + "&l=" + location)
r = requests.get(url)
rcontent = r.content
prettify = BeautifulSoup(rcontent, "html.parser")
all_job_url = []
for tag in prettify.find_all('div', {'data-tn-element':"jobTitle"}):
for links in tag.find_all('a'):
print (links['href'])
You should be looking for the anchor a tag. It looks like this:
<a class="turnstileLink" data-tn-element="jobTitle" href="/rc/clk?jk=3611ac98c0167102&fccid=459dce363200e1be" ...>Project <b>Engineer</b></a>
Call soup.find_all and iterate over the result set, extracting the links through the href attribute.
import requests
from bs4 import BeautifulSoup
# valid query, replace with something else
url = "https://ca.indeed.com/jobs?q=engineer&l=Calgary%2C+AB"
r = requests.get(url)
soup = BeautifulSoup(r.content, "html.parser")
all_job_url = []
for tag in soup.find_all('a', {'data-tn-element':"jobTitle"}):
all_job_url.append(tag['href'])

Beautifulsoup to retrieve the href list

Thanks for attention!
I'm trying to retrieve the href of products in search result.
For example this page:
However When I narrow down to the product image class, the retrived href are image links....
Can anyone solve that? Thanks in advance!
url = 'http://www.homedepot.com/b/Husky/N-5yc1vZrd/Ntk-All/Ntt-chest%2Band%2Bcabinet?Ntx=mode+matchall&NCNI-5'
content = urllib2.urlopen(url).read()
content = preprocess_yelp_page(content)
soup = BeautifulSoup(content)
content = soup.findAll('div',{'class':'content dynamic'})
draft = str(content)
soup = BeautifulSoup(draft)
items = soup.findAll('div',{'class':'cell_section1'})
draft = str(items)
soup = BeautifulSoup(draft)
content = soup.findAll('div',{'class':'product-image'})
draft = str(content)
soup = BeautifulSoup(draft)
You don't need to load the content of each found tag with BeautifulSoup over and over again.
Use CSS selectors to get all product links (a tag under a div with class="product-image")
import urllib2
from bs4 import BeautifulSoup
url = 'http://www.homedepot.com/b/Husky/N-5yc1vZrd/Ntk-All/Ntt-chest%2Band%2Bcabinet?Ntx=mode+matchall&NCNI-5'
soup = BeautifulSoup(urllib2.urlopen(url))
for link in soup.select('div.product-image > a:nth-of-type(1)'):
print link.get('href')
Prints:
http://www.homedepot.com/p/Husky-41-in-16-Drawer-Tool-Chest-and-Cabinet-Set-HOTC4016B1QES/205080371
http://www.homedepot.com/p/Husky-26-in-6-Drawer-Chest-and-Cabinet-Combo-Black-C-296BF16/203420937
http://www.homedepot.com/p/Husky-52-in-18-Drawer-Tool-Chest-and-Cabinet-Set-Black-HOTC5218B1QES/204825971
http://www.homedepot.com/p/Husky-26-in-4-Drawer-All-Black-Tool-Cabinet-H4TR2R/204648170
...
div.product-image > a:nth-of-type(1) CSS selector would match every first a tag directly under the div with class product-image.
To save the links into a list, use a list comprehension:
links = [link.get('href') for link in soup.select('div.product-image > a:nth-of-type(1)')]

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