CODEWITHSUNDEEP
pythonseleniumselenium-webdriverweb-scraping
Xonshiz
Xonshiz

Reputation: 1367

Fetch all href link using selenium in python

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)

Upvotes: 55

Views: 158722

Answers (11)

srinivas s
srinivas s

Reputation: 81

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.

Upvotes: 5

JRodDynamite
JRodDynamite

Reputation: 12613

For Selenium >4.3.0, you can try the following:

from selenium.webdriver.common.by import By

elems = driver.find_elements(by=By.XPATH, "//a[@href]")
for elem in elems:
    print(elem.get_attribute("href"))

You can also refer to answers here for a more detailed explanation.

Note: Below solution works for Selenium <4.3.0.

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).

Upvotes: 107

Chris
Chris

Reputation: 18876

For 2023:

url = "https://example.com"
driver.get(url)
raw_links = driver.find_elements(By.XPATH, '//a [@href]')
for link in raw_links:
    l = link.get_attribute("href")
    print("raw_link:{}".format(l))

Upvotes: 1

DataMinion
DataMinion

Reputation: 437

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

Upvotes: 3

Johannes Stephan
Johannes Stephan

Reputation: 65

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"))

Upvotes: 0

Suman Das
Suman Das

Reputation: 11

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')

Upvotes: 1

emehex
emehex

Reputation: 10528

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]

Upvotes: 2

Gabriel Chung
Gabriel Chung

Reputation: 1577

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)

Upvotes: 10

Anupriya Nishad
Anupriya Nishad

Reputation: 21

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)

Upvotes: 0

Shawn
Shawn

Reputation: 611

You can try something like:

    links = driver.find_elements_by_partial_link_text('')

Upvotes: 3

Python_Novice
Python_Novice

Reputation: 170

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

Upvotes: 2

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