After thorough investigation and experimentation, I have confirmed the existence of a function called find_elements_by_tag_name() that you can employ. In my own testing, this particular example performed without any issues.

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: It's crucial to include a delay in the script. Run it in debug mode first to ensure the URL page loads successfully. If the page loading is slow, adjust the delay (sleep time) accordingly before extracting the data.

If you encounter any issues, feel free to check out the explanation provided in the link below or leave a comment.

Learn how to extract links from a webpage using selenium webdriver

Here is a possible solution:

    Consider using the following code snippet:

links = driver.find_elements_by_partial_link_text('')

The traditional way of using Selenium's driver.find_elements_by_*** is no longer compatible with Selenium 4. The current recommended approach is to utilize find_elements() along with the By class.

Approach 1: For loop

This method involves two lists, one for By.XPATH and the other for By.TAG_NAME, each serving a unique purpose. However, you can choose to use either of them as they are not both necessary.

Personally, I find By.XPATH more straightforward since it avoids returning a seemingly irrelevant None value compared to By.TAG_NAME. Additionally, this code snippet eliminates duplicates in the results.

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:
    link = elem.get_attribute("href")
    if link not in href_links:
        href_links.append(link)

for elem in elems2:
    link = elem.get_attribute("href")
    if (link not in href_links2) & (link is not None):
        href_links2.append(link)

print(len(href_links))  # 360
print(len(href_links2))  # 360

print(href_links == href_links2)  # True

Approach 2: List Comprehension

If duplicate links are acceptable, a single line list comprehension can be employed.

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

To incorporate the HTML dom into your Python project, you can utilize the html dom library. You can easily access and install this library using PIP by visiting the following link:

https://pypi.python.org/pypi/htmldom/2.0

from htmldom import htmldom
dom = htmldom.HtmlDom("https://www.github.com/")  
dom = dom.createDom()

The provided code snippet initializes a HtmlDom object. The HtmlDom requires a default parameter, which is the URL of the page you want to parse. After creating the dom object, you need to execute the "createDom" method of HtmlDom. This will analyze the HTML content and generate a parse tree that can be utilized for searching and modifying the HTML data. The only requirement imposed by the library is that the data (whether it's HTML or XML) must have a root element.

You can retrieve elements using the "find" method of the HtmlDom object:

p_links = dom.find("a")  
for link in p_links:
  print ("URL: " +link.attr("href"))

This piece of code will display all the links/URLs present on the webpage

Unfortunately, it seems that the original link provided by OP is no longer working...

If you're interested in extracting links from a webpage, here's a method to fetch all the "Hot Network Questions" links from this page using 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]

If you want to extract links from a webpage, you can easily achieve this using BeautifulSoup in a straightforward and effective manner. I have personally tested the code snippet below and it worked flawlessly for the same purpose.

Simply add the following lines of code after the line:

driver.get("http://example.com/")

Insert the code snippet below:

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')

Looking ahead to 2023:

target_url = "https://example.com"
driver.get(target_url)
all_links = driver.find_elements(By.XPATH, '//a [@href]')
for ind_link in all_links:
    current_link = ind_link.get_attribute("href")
    print("Found link:{}".format(current_link))

import requests
from selenium import webdriver
import bs4
driver = webdriver.Chrome(r'C:\chromedrivers\chromedriver') #specify the filepath for Chrome driver
data=requests.get('https://google.co.in/') #choose any website you want to scrape
soup=bs4.BeautifulSoup(data.text,'html.parser')
for link in soup.findAll('a'):
    print(link)

Revision regarding the previous answer:

elements = driver.find_elements_by_xpath("//a[@href]")
for element in elements:
    print(element.get_attribute("href"))