My code goes into a website and scrapes rows of information (title and time).
However, there is one tag ('p') that I am not sure how to get using 'get element by'.
On the website, it is the information under each title.
Here is my code so far:
import time
from selenium import webdriver
from bs4 import BeautifulSoup
import requests
driver = webdriver.Chrome()
driver.get('https://www.nutritioncare.org/ASPEN21Schedule/#tab03_19')
driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
eachRow = driver.find_elements_by_class_name('timeline__item')
time.sleep(1)
for item in eachRow:
time.sleep(1)
title = item.find_element_by_class_name('timeline__item-title')
tim = item.find_element_by_class_name('timeline__item-time')
tex = item.find_element_by_tag_name('p') # This is the part I don’t know how to scrape
print(title.text, tim.text, tex.text)
I checked the page and there are several p tags, I suggest to use find_elements_by_tag_name instead of find_element_by_tag_name (to get all the p tags including the p tag that you want) and iterate over all the p tags elements and then join the text content and do strip on it.
from selenium import webdriver
from bs4 import BeautifulSoup
import time
import requests
driver = webdriver.Chrome()
driver.get('https://www.nutritioncare.org/ASPEN21Schedule/#tab03_19')
driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
eachRow = driver.find_elements_by_class_name('timeline__item')
time.sleep(1)
for item in eachRow:
time.sleep(1)
title=item.find_element_by_class_name('timeline__item-title')
tim=item.find_element_by_class_name('timeline__item-time')
tex=item.find_elements_by_tag_name('p')
text = " ".join([i.text for i in tex]).strip()
print(title.text,tim.text, text)
Since the webpage has several p tags, it would be better to use the .find_elements_by_class() method. Replace the print call in the code with the following:
print(title.text,tim.text)
for t in tex:
if t.text == '':
continue
print(t.text)
Maybe try using different find_elements_by_class... I don't use Python that much, but try this unless you already have.
I have watched a video at this link https://www.youtube.com/watch?v=EELySnTPeyw and this is the code ( I have changed the xpath as it seems the website has been changed)
import selenium.webdriver as webdriver
def get_results(search_term):
url = 'https://www.startpage.com'
browser = webdriver.Chrome(executable_path="D:\\webdrivers\\chromedriver.exe")
browser.get(url)
search_box = browser.find_element_by_id('q')
search_box.send_keys(search_term)
try:
links = browser.find_elements_by_xpath("//a[contains(#class, 'w-gl__result-title')]")
except:
links = browser.find_lemets_by_xpath("//h3//a")
print(links)
for link in links:
href = link.get_attribute('href')
print(href)
results.append(href)
browser.close()
get_results('cat')
The code works well as for the part of opening the browser and navigating to the search box and sending keys but as for the links return an empty list although I have manually searched for the xpath in the developer tools and it returns 10 results.
You need to add keys.enter to your search. You weren't on the next page.
search_box.send_keys(search_term+Keys.ENTER)
Import
from selenium.webdriver.common.keys import Keys
Outputs
https://en.wikipedia.org/wiki/Cat
https://www.cat.com/en_US.html
https://www.cat.com/
https://www.youtube.com/watch?v=cbP2N1BQdYc
https://icatcare.org/advice/thinking-of-getting-a-cat/
https://www.caterpillar.com/en/brands/cat.html
https://www.petfinder.com/cats/
https://www.catfootwear.com/US/en/home
https://www.aspca.org/pet-care/cat-care/general-cat-care
https://www.britannica.com/animal/cat
My code so far - If I search for a job title in LinkedIn - (For example-Cyber Analyst), will gather all links of this job posting/page
Goal -I put these links in a list, and iterate through them (Code works so far) to print the title of each job posting/link
My code iterates through every link, but does not get the Post title/Job title text. Which is the goal.
import time
from selenium import webdriver
from selenium.webdriver.chrome.options import Options
from webdriver_manager.chrome import ChromeDriverManager
test1=[]
options = Options()
options.headless = True
driver = webdriver.Chrome(ChromeDriverManager().install())
url = "https://www.linkedin.com/jobs/search/?currentJobId=2213597199&geoId=103644278&keywords=cyber%20analyst&location=United%20States&start=0&redirect=false"
driver.get(url)
time.sleep(2)
elements = driver.find_elements_by_class_name("result-card__full-card-link")
job_links = [e.get_attribute("href") for e in elements]
for job_link in job_links:
test1.append(job_link) #prints all links into test1
for b in test1:
driver.get(b)
time.sleep(3)
element1=driver.find_elements_by_class_name("jobs-top-card__job-title t-24")
title=[t.get_attribute("jobs-top-card__job-title t-24") for t in element1]
print(title)
I couldn't see class 'obs-top-card__job-title t-24' on the link pages, but this gives you the job titles for every href
Change
element1=driver.find_elements_by_class_name("jobs-top-card__job-title t-24")
title=[t.get_attribute("jobs-top-card__job-title t-24") for t in element1]
to
element1=driver.find_elements_by_class_name("topcard__title")
title=[t.text for t in element1]
>>> ['Cyber Threat Intelligence Analyst']
>>> ['Jr. Python/Cyber Analyst (TS/SCI)']
>>> ['Cyber Security Analyst']
....ect
every time you do driver.get(b) a new page is fetched, so the html code is not the same as driver.get(url) so I think t.get_attribute("jobs-top-card__job-title t-24") belongs to html code for driver.get(url) but as I said this page is closed as driver.get(b) is fetched
Also each page for driver.get(b) has the same structure so element1=driver.find_elements_by_class_name("topcard__title") will always work
e.g. this is a one of the pages of driver.get(b):
This is where topcard_title is
This question already has an answer here:
How to click a random link from google search results through Selenium and Python
(1 answer)
Closed 4 years ago.
I'm aiming to get URLs as a results for my search_term. By running the below code in Python, my only output is []. Would someone be able to help to modify the code in order to receive the list with URLs to search results, ideally limited to first 10-20 URLs? Thanks in advance, please find the code below:
import selenium.webdriver as webdriver
def get_results(search_term):
url = "https://www.google.com"
browser = webdriver.Safari()
browser.get(url)
search_box = browser.find_element_by_name("q")
search_box.send_keys(search_term)
search_box.submit()
try:
links = browser.find_elements_by_xpath("//ol[#class='web_regular_results']//h3//a")
except:
links = browser.find_elements_by_xpath("//h3//a")
results = []
for link in links:
href = link.get_attribute("href")
print(href)
results.append(href)
browser.close()
return results
get_results("fish")
You have provided wrong xpath that is why you are not getting it.I have tried with chrome and change the xpath it is working fine.Try this and let know.
def get_results(search_term):
url = "https://www.google.com"
browser = webdriver.Safari()
browser.get(url)
search_box = browser.find_element_by_name("q")
search_box.send_keys(search_term)
search_box.submit()
try:
links = browser.find_elements_by_xpath("//h3[#class='r']/a")
except:
links = browser.find_elements_by_xpath("//h3[#class='r']/descendant::a")
print(len(links))
results = []
for link in links:
href = link.get_attribute("href")
print(href)
results.append(href)
browser.close()
return results
get_results("fish")
I am practicing Selenium in Python and I wanted to fetch all the links on a web page using Selenium.
For example, I want all the links in the href= property of all the <a> tags on http://psychoticelites.com/
I've written a script and it is working. But, it's giving me the object address. I've tried using the id tag to get the value, but, it doesn't work.
My current script:
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
driver = webdriver.Firefox()
driver.get("http://psychoticelites.com/")
assert "Psychotic" in driver.title
continue_link = driver.find_element_by_tag_name('a')
elem = driver.find_elements_by_xpath("//*[#href]")
#x = str(continue_link)
#print(continue_link)
print(elem)
Well, you have to simply loop through the list:
elems = driver.find_elements_by_xpath("//a[#href]")
for elem in elems:
print(elem.get_attribute("href"))
find_elements_by_* returns a list of elements (note the spelling of 'elements'). Loop through the list, take each element and fetch the required attribute value you want from it (in this case href).
I have checked and tested that there is a function named find_elements_by_tag_name() you can use. This example works fine for me.
elems = driver.find_elements_by_tag_name('a')
for elem in elems:
href = elem.get_attribute('href')
if href is not None:
print(href)
driver.get(URL)
time.sleep(7)
elems = driver.find_elements_by_xpath("//a[#href]")
for elem in elems:
print(elem.get_attribute("href"))
driver.close()
Note: Adding delay is very important. First run it in debug mode and Make sure your URL page is getting loaded. If the page is loading slowly, increase delay (sleep time) and then extract.
If you still face any issues, please refer below link (explained with an example) or comment
Extract links from webpage using selenium webdriver
You can try something like:
links = driver.find_elements_by_partial_link_text('')
You can import the HTML dom using html dom library in python. You can find it over here and install it using PIP:
https://pypi.python.org/pypi/htmldom/2.0
from htmldom import htmldom
dom = htmldom.HtmlDom("https://www.github.com/")
dom = dom.createDom()
The above code creates a HtmlDom object.The HtmlDom takes a default parameter, the url of the page. Once the dom object is created, you need to call "createDom" method of HtmlDom. This will parse the html data and constructs the parse tree which then can be used for searching and manipulating the html data. The only restriction the library imposes is that the data whether it is html or xml must have a root element.
You can query the elements using the "find" method of HtmlDom object:
p_links = dom.find("a")
for link in p_links:
print ("URL: " +link.attr("href"))
The above code will print all the links/urls present on the web page
Unfortunately, the original link posted by OP is dead...
If you're looking for a way to scrape links on a page, here's how you can scrape all of the "Hot Network Questions" links on this page with gazpacho:
from gazpacho import Soup
url = "https://stackoverflow.com/q/34759787/3731467"
soup = Soup.get(url)
a_tags = soup.find("div", {"id": "hot-network-questions"}).find("a")
[a.attrs["href"] for a in a_tags]
You can do this by using BeautifulSoup with very easy and efficient way. I have tested the below codes and worked fine for the same purpose.
After this line -
driver.get("http://psychoticelites.com/")
use the below code -
response = requests.get(browser.current_url)
soup = BeautifulSoup(response.content, 'html.parser')
for link in soup.find_all('a'):
if link.get('href'):
print(link.get("href"))
print('\n')
All of the accepted answers using Selenium's driver.find_elements_by_*** no longer work with Selenium 4. The current method is to use find_elements() with the By class.
Method 1: For loop
The below code utilizes 2 lists. One for By.XPATH and the other, By.TAG_NAME. One can use either-or. Both are not needed.
By.XPATH IMO is the easiest as it does not return a seemingly useless None value like By.TAG_NAME does. The code also removes duplicates.
from selenium.webdriver.common.by import By
driver.get("https://www.amazon.com/")
href_links = []
href_links2 = []
elems = driver.find_elements(by=By.XPATH, value="//a[#href]")
elems2 = driver.find_elements(by=By.TAG_NAME, value="a")
for elem in elems:
l = elem.get_attribute("href")
if l not in href_links:
href_links.append(l)
for elem in elems2:
l = elem.get_attribute("href")
if (l not in href_links2) & (l is not None):
href_links2.append(l)
print(len(href_links)) # 360
print(len(href_links2)) # 360
print(href_links == href_links2) # True
Method 2: List Comprehention
If duplicates are OK, one liner list comprehension can be used.
from selenium.webdriver.common.by import By
driver.get("https://www.amazon.com/")
elems = driver.find_elements(by=By.XPATH, value="//a[#href]")
href_links = [e.get_attribute("href") for e in elems]
elems2 = driver.find_elements(by=By.TAG_NAME, value="a")
# href_links2 = [e.get_attribute("href") for e in elems2] # Does not remove None values
href_links2 = [e.get_attribute("href") for e in elems2 if e.get_attribute("href") is not None]
print(len(href_links)) # 387
print(len(href_links2)) # 387
print(href_links == href_links2) # True
import requests
from selenium import webdriver
import bs4
driver = webdriver.Chrome(r'C:\chromedrivers\chromedriver') #enter the path
data=requests.request('get','https://google.co.in/') #any website
s=bs4.BeautifulSoup(data.text,'html.parser')
for link in s.findAll('a'):
print(link)
Update for the existing solving Post:
For the current version it needs to be:
elems = driver.find_elements_by_xpath("//a[#href]")
for elem in elems:
print(elem.get_attribute("href"))