Get Current Url From Browser Using Python

I am running an HTTP server which serves a bitmap according to the dimensions in the browser URL i.e localhost://image_x120_y30.bmp. My server is running in infinite loop and I wan

Solution 1:

If to use Selenium for web navigation:

from selenium import webdriver
driver = webdriver.Firefox()
print (driver.current_url)

Solution 2:

You can get the current url by doing path_info = request.META.get('PATH_INFO') http_host = request.META.get('HTTP_HOST'). You can add these two to get complete url. Basically request.META returns you a dictionary which contain a lot of information. You can try it.

Solution 3:

I just solved a class problem similar to this. We've been using Splinter to walk through pages (you will need to download splinter and Selenium). As I walk through pages, I periodically need to pull the url of the page I'm currently on. I do that using the command new_url = browser.url Below is an example of my code.

I do this using the following code.

##import dependenciesfrom splinter import browser
import requests


## go to original page 
browser.visit(url)

## Loop through the page associated with each headlinefor headline in titles:
    print(headline.text)
    browser.click_link_by_partial_text(headline.text)
## Now that I'm on the new page, I need to grab the url
    new_url = browser.url
    print(new_url)
## Go back to original page
    browser.visit(url)

Solution 4:

Below is the solution I use in Django.

For eg.,. if browser url is https://www.example.com/dashboard

try:
    from urlparse import urlparse
except ImportError:
    from urllib.parse import urlparse

frontend_url = request.META.get('HTTP_REFERER')
url = urlparse(frontend_url)
print (url)
# ParseResult(scheme='https', netloc='example.com', path='/dashboard', params='', query='', fragment='')

Solution 5:

You could use the requests module:

import requests


link = "https://stackoverflow.com"data = requests.request("GET", link)
url = data.url