Fetch all href link using selenium in python

Question

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= prop

Answer 1

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]

Answer 2

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

Answer 3

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)

Answer 4

You can try something like:

    links = driver.find_elements_by_partial_link_text('')

Answer 5

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).

Answer 6

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)