title: How to get the data of a website's "pop-up" box with Selenium Webdriver in Python 3
body: <p>I'm a newcomer in data scraping, so I appologize in advance if my question is flawed by any reason.</p>
  
  <p>I'm trying to scrap a page of an air company (<a href="https://www.voegol.com.br/en-us/Paginas/Default.aspx" rel="nofollow">link</a>, as to get the data of the flight (ex: aircraft type). I was successful as to enter the data for the flights (departure and arrival airports / data), and get a screen with the flight proposals. </p>
  
  <p>In the second screen, there is a link called "Direct flight". I also included a code line to click on this, so a new pop-up window appears - this one, with the data I want (scheduled departure/arrival times, aircraft type).</p>
  
  <p>But when I try to download it with "html = browser.page_source" (parsing with BeautifulSoup), it apparently is downloading only the content of the previous page (before clicking on "Direct flight", while I want to pick the information from the pop-up box (see <a href="http://i.stack.imgur.com/A31Hi.jpg" rel="nofollow">screenshots</a>).</p>
  
  <pre><code>voosdiretos=browser.find_elements_by_class_name('plusBus')
  voo=voosdiretos[0]
  voo.click()
  
  html = browser.page_source
  soup = BeautifulSoup(html)
  
  soup_string=str(soup)
  print('soup_to_string')
  </code></pre>
  
  <p>I tried to find a solution for it. Those usually recommended the use of window_handle, but there is no way I can make it work here (I suspect this pop-up windows is not an actual new pop-up, but some sort of javascript pop-up window).</p>
  
  <p>Does anybody have any suggestions on how to scrap this information?</p>
  
  <p>[EDIT]</p>
  
  <p>Following Grasshopper's suggestion, I tried to get the elements:</p>
  
  <pre><code>elem_=browser.find_elements_by_css_selector('.informacoesLightbox bgGrid borderIe8')
  print(len(elem_))
  print(type(elem_))
  </code></pre>
  
  <p>Output was 0 and  (nothing was returned).</p>
  
  <p>Any suggestion?</p>