Safely open a list of links in Firefox

Question

I receive a list of links via http and I want to open them all in a new Firefox (or default browser) window as tabs; that is my script must not touch existing windows.

This is the best solution I could come up with:

from subprocess import call
links = GetListViaHTTP()
if len(links) == 1:
 call(['firefox', '-new-window'] + links)
elif len(links) > 1:
 call(['firefox'] + links)
else:
 print('Nothing to do.')

This worked like a charm when I tested it, but putting something I received via http into subprocess.call does not look save to me at all. So the question is:

How can I improve this code to avoid being vulnerable to MITM attacks?

If the list was not tampered with by an attacker, all links it contains will point to Stack Overflow answers, so maybe somehow validating the links with urllib.parse is good enough?

• Peilonrayz: Does the program have to open the link in firefox? Or can you open it in the default web-browser?
• Baum mit Augen: @Peilonrayz Default webbrowser is fine too if it makes things easier. (It's technically even preferred, but I wanted to avoid my question being too broad. I'll gladly take either.)

Answer 1

Rather than implementing this yourself, Python by default has webbrowser. This allows you to open one link in a; new tab, or new window. And so, to open all links in the list with the default web-browser you can:

import webbrowser
for link in links:
 webbrowser.open_new_tab(link)

Since you say you don't trust the source of these urls, you can make sure they are actually URLs, and check there is no shell injection going on. This question says how you could do that.

And so I personally would use:

import webbrowser
import rfc3987
def open_links(links):
 links = (
 link
 for link in links
 if rfc3987.match(link, rule='IRI') is not None
 )
 # Get default webbrowser, change if you want to force a specific browser
 browser = webbrowser.get()
 try:
 browser.open_new(next(links))
 except StopIteration:
 pass
 else:
 for link in links:
 browser.open_new_tab(link)

• Baum mit Augen: Unfortunately, open_new opens new tabs in existing windows on my system (Arch Linux). I don't know why, maybe I botched my setup. But that rfc3987 package seems really useful to start with, thank you. (Btw, webbrowser.open_tab does not seem to exist, I guess that's a typo?)
• Peilonrayz: @BaummitAugen That's odd, it may just not be supported. But that was a typo, I'll fix it now, :)
• Baum mit Augen: Addendum: When I turn off the "Open new windows in new tabs" option in firefox, I get a new tab for every link. Probably Firefox' fault?
• Baum mit Augen: Alright, after a little bit more research it appears to be the Browser's fault that open_new and open_new_tab are not handled properly. But validating the links with the rfc3987 package sounds save enough (hope I am right :) ), so that counts as the answer. Thanks for your time!

Answer 2

You can also control a real browser with Python selenium bindings:

pip install selenium

Then, you can combine it with @Peilonrayz's link checking code:

from selenium import webdriver
# ...
driver = webdriver.Firefox()
try:
 driver.get(next(links))
except StopIteration:
 pass
else:
 for link in links:
 body = driver.find_element_by_tag_name("body")
 body.send_keys(Keys.CONTROL + 't') # opens a new tab
 driver.get(link)

(please replace CONTROL with COMMAND on Mac OS).

• Baum mit Augen: Thanks, I tried that too, but it is extremely slow and looks very clumsy compared to the CLI variant with call.
• alecxe: @BaummitAugen well, yes, there is a substantial overhead, but, if you would need to do something else except opening the links in the browser, webbrowser or a system call would not be an option anymore. For example, if you would need to click a specific element, or send a text to an input. But, if all you need is to open links in the browser, then yes, selenium is too much overhead. Thanks.