ruby selenium webkit capybara capybara-webkit

Get image file using ruby & capybara

It is image tag on the page accessed by capybara via HTTPS protocol:

<img src="path">

Is it any way to get image file from the page using capybara with any kind of driver?

I can not use something like File.read('path') because image is also accessible via HTTPS only. My latest researches brought me to such kind of solution:

Visit page
Save page to png (webkit driver has such useful ability)
Crop image

But I do believe that pretty solution exists.

Edited 1:

I've tried out padde's solution, but here is response body:

<html><head><title>Object moved</title></head> 
    <body>
        <h2>Object moved to <a href=\"/Bledy/Blad404.aspx?aspxerrorpath=/CaptchaType.ashx\">here</a>.</h2> 
    </body>
</html>

Edited 2:

> curl -I image_path

5860cf30abf5d5480
HTTP/1.1 302 Found
Cache-Control: private
Content-Length: 168
Content-Type: text/html; charset=utf-8
Location: /Bledy/Blad404.aspx?aspxerrorpath=/CaptchaType.ashx
Server: Microsoft-IIS/7.5
X-AspNet-Version: 4.0.30319
X-Powered-By: ASP.NET
Date: Sat, 03 Nov 2012 17:18:55 GMT

Solution

What you probably want is a HTTPS request from Ruby if i get this right. Try:

require 'net/https'

url = URI.parse('path')

Net::HTTP.start(url.host, url.port, :use_ssl => true, :verify_mode => OpenSSL::SSL::VERIFY_NONE) do |http|
  res = http.get(url.request_uri)
  open("image.png", "wb") do |f|
    f.write(res.body)
  end
end

For cropping, you can either use chunky_png (pure Ruby) or rmagick (requires ImageMagick)

Edit: If you want to follow redirects you can do

require 'net/https'

def process_image( content )
  # do your cropping here

  open("image.png", "wb") do |f|
    f.write(content)
  end
end

def fetch( url )
  Net::HTTP.start(url.host, url.port, :use_ssl => true, :verify_mode => OpenSSL::SSL::VERIFY_NONE) do |http|
    response = http.get(url.request_uri)
    case response.code
    when Net::HTTPRedirection
      fetch response['location']
    else
      process_image response.body
    end
  end
end

fetch URI.parse('path')