Login to a website through web-scraping tool in Python

Question

I am using Selenium webdriver in Python for a web-scraping project.

I would like to login by entering the login details and then click the submit button.

I am able to enter the Username and Password. But I am not able to mouseclick the submit button.

The "submit" button is of type <input>.

<input type="image" src="/images/buttons/loginnow.gif" tabindex="3">

Here is the python code where I am trying to click the mouse.

submitButton=driver.find_element_by_xpath("//input[@type='image'][@src='/images/buttons/loginnow.gif']")
driver.click(submitButton)

I get the following error :

AttributeError: 'WebDriver' object has no attribute 'click'

Any idea how to fix it or any other alternative solution to login to a website in Python.

Thanks

Answer 1

I had good luck using mechanize. It's pretty straightforward and simple to use.

Here's a stripped-down version of a script I made:

from BeautifulSoup import BeautifulSoup
from tidylib import tidy_document

import mechanize
import cookielib

if __name__ == '__main__':
  browser = mechanize.Browser()

  cookiejar = cookielib.LWPCookieJar()
  browser.set_cookiejar(cookiejar)

  browser.set_handle_equiv(True)
  browser.set_handle_redirect(True)
  browser.set_handle_referer(True)
  browser.set_handle_robots(False)

  browser.open('https://www.example.com/')

  browser.select_form(name = 'loginform')
  browser['username'] = 'foo'
  browser['password'] = 'bar'

  browser.submit()

  browser.open(browser.click_link(text = 'Link text'))

  soup = BeautifulSoup(tidy_document(browser.response().read())[0])

You don't need to click on the image, really. You just need to fill out all the appropriate form details and just submit() it.

Also, if you won't be parsing anything, just get rid of the BeautifulSoup and tidylib dependencies.

Answer 2

You need to call the click function of the element, not the driver.

submitButton=driver.find_element_by_xpath("//input[@type='image'][@src='/images/buttons/loginnow.gif']")
submitButton.click()